Hibbeler dynamics ch12

Although each of these planes is rather large, from a distance their motioncan be modeled as if each plane were a particle

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3

CHAPTER OBJECTIVES

Kinematics of a Particle

• To introduce the concepts of position, displacement, velocity, andacceleration.

• To study particle motion along a straight line and represent thismotion graphically.

• To investigate particle motion along a curved path using differentcoordinate systems.

• To present an analysis of dependent motion of two particles.

• To examine the principles of relative motion of two particles usingtranslating axes.

C H A P T E R

12

12.1 Introduction

Mechanics is a branch of the physical sciences that is concernedwith the state of rest or motion of bodies subjected to theaction of forces. The mechanics of rigid bodies is divided intotwo areas: statics and dynamics. Statics is concerned with theequilibrium of a body that is either at rest or moves with constantvelocity. The foregoing treatment is concerned with dynamicswhich deals with the accelerated motion of a body. Here thesubject of dynamics will be presented in two parts: kinematics,which treats only the geometric aspects of the motion, andkinetics, which is the analysis of the forces causing the motion.To develop these principles, the dynamics of a particle will bediscussed first, followed by topics in rigid-body dynamics intwo and then three dimensions.

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4 • CHAPTER 12 Kinematics of a Particle

Historically, the principles of dynamics developed when it was possibleto make an accurate measurement of time. Galileo Galilei (1564–1642)was one of the first major contributors to this field. His work consistedof experiments using pendulums and falling bodies. The most significantcontributions in dynamics, however, were made by Isaac Newton(1642–1727), who is noted for his formulation of the three fundamentallaws of motion and the law of universal gravitational attraction. Shortlyafter these laws were postulated, important techniques for theirapplication were developed by Euler, D’Alembert, Lagrange, and others.

There are many problems in engineering whose solutions requireapplication of the principles of dynamics. Typically the structural designof any vehicle, such as an automobile or airplane, requires considerationof the motion to which it is subjected. This is also true for manymechanical devices, such as motors, pumps, movable tools, industrialmanipulators, and machinery. Furthermore, predictions of the motionsof artificial satellites, projectiles, and spacecraft are based on the theoryof dynamics. With further advances in technology, there will be an evengreater need for knowing how to apply the principles of this subject.

Problem Solving. Dynamics is considered to be more involved thanstatics since both the forces applied to a body and its motion must betaken into account.Also, many applications require using calculus, ratherthan just algebra and trigonometry. In any case, the most effective way oflearning the principles of dynamics is to solve problems. To be successfulat this, it is necessary to present the work in a logical and orderly manneras suggested by the following sequence of steps:

1. Read the problem carefully and try to correlate the actual physicalsituation with the theory studied.

2. Draw any necessary diagrams and tabulate the problem data.

3. Establish a coordinate system and apply the relevant principles,generally in mathematical form.

4. Solve the necessary equations algebraically as far as practical; then,use a consistent set of units and complete the solution numerically.Report the answer with no more significant figures than theaccuracy of the given data.

5. Study the answer using technical judgment and common sense todetermine whether or not it seems reasonable.

6. Once the solution has been completed, review the problem. Try tothink of other ways of obtaining the same solution.

In applying this general procedure, do the work as neatly as possible.Being neat generally stimulates clear and orderly thinking, and vice versa.

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SECTION 12.2 Rectilinear Kinematics: Continuous Motion • 5

12.2 Rectilinear Kinematics: Continuous Motion

We will begin our study of dynamics by discussing the kinematics of aparticle that moves along a rectilinear or straight line path. Recall thata particle has a mass but negligible size and shape. Therefore we mustlimit application to those objects that have dimensions that are of noconsequence in the analysis of the motion. In most problems, one isinterested in bodies of finite size, such as rockets, projectiles, or vehicles.Such objects may be considered as particles, provided motion of the bodyis characterized by motion of its mass center and any rotation of thebody is neglected.

Rectilinear Kinematics. The kinematics of a particle is characterizedby specifying, at any given instant, the particle’s position, velocity, andacceleration.

Position. The straight-line path of a particle will be defined using asingle coordinate axis s, Fig. 12–1a. The origin O on the path is a fixedpoint, and from this point the position vector r is used to specify thelocation of the particle P at any given instant. Notice that r is always alongthe s axis, and so its direction never changes. What will change is itsmagnitude and its sense or arrowhead direction. For analytical work it istherefore convenient to represent r by an algebraic scalar s, representingthe position coordinate of the particle, Fig. 12–1a.The magnitude of s (andr) is the distance from O to P, usually measured in meters (m) or feet (ft),and the sense (or arrowhead direction of r) is defined by the algebraicsign on s.Although the choice is arbitrary, in this case s is positive since thecoordinate axis is positive to the right of the origin. Likewise, it is negativeif the particle is located to the left of O.

Displacement. The displacement of the particle is defined as the changein its position. For example, if the particle moves from P to Fig. 12–1b,the displacement is Using algebraic scalars to represent we also have

Here is positive since the particle’s final position is to the right of itsinitial position, i.e., Likewise, if the final position were to the leftof its initial position, would be negative.

Since the displacement of a particle is a vector quantity, it should bedistinguished from the distance the particle travels. Specifically, thedistance traveled is a positive scalar which represents the total length ofpath over which the particle travels.

¢ss¿ 7 s.

¢s

¢s = s¿ - s

¢r,¢r = r¿ - r.P¿,

r

s

sP

Position!

(a)

O

Fig. 12–1A

r

s

sP

Displacement!

(b)

s'

P'

O

r'

∆s

∆r

Fig. 12–1

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Velocity. If the particle moves through a displacement from P to during the time interval Fig. 12–1b, the average velocity of the particleduring this time interval is

If we take smaller and smaller values of the magnitude of becomessmaller and smaller. Consequently, the instantaneous velocity is definedas or

Representing v as an algebraic scalar, Fig. 12–1c, we can also write

(12–1)

Since or dt is always positive, the sign used to define the sense of thevelocity is the same as that of or ds. For example, if the particle is movingto the right, Fig. 12–1c, the velocity is positive; whereas if it is moving to theleft, the velocity is negative. (This is emphasized here by the arrow writtenat the left of Eq. 12–1.) The magnitude of the velocity is known as the speed,and it is generally expressed in units of m s or ft s.

Occasionally, the term “average speed” is used. The average speed isalways a positive scalar and is defined as the total distance traveled bya particle, divided by the elapsed time i.e.,

For example the particle in Fig. 12–1d travels along the path of lengthin time so its average speed is but its average

velocity is vavg = - ¢s>¢t.1vsp2avg = sT>¢t,¢t,sT

1vsp2avg =sT

¢t

¢t;sT,

>>¢s

¢t

v =ds

dt1:+ 2

v =drdt

v = lim¢t:0

1¢r>¢t2, ¢r¢t,

vavg =¢r¢t

¢t,P¿¢r

sP

Velocity!

(c)

O∆s

P'

v

Fig. 12–1C

–∆s

sP

sT

Average velocity and!Average speed!

(d)

O

P'

Fig. 12–1

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Acceleration. Provided the velocity of the particle is known at the twopoints P and the average acceleration of the particle during the timeinterval is defined as

Here represents the difference in the velocity during the time intervali.e., Fig. 12–1e.

The instantaneous acceleration at time t is found by taking smaller andsmaller values of and corresponding smaller and smaller values of

so that or, using algebraic scalars,

(12–2)

Substituting Eq. 12–1 into this result, we can also write

Both the average and instantaneous acceleration can be either positiveor negative. In particular, when the particle is slowing down, or its speedis decreasing, it is said to be decelerating. In this case, in Fig. 12–1f isless than and so will be negative. Consequently, a willalso be negative, and therefore it will act to the left, in the opposite senseto v. Also, note that when the velocity is constant, the acceleration is zerosince Units commonly used to express the magnitudeof acceleration are or

A differential relation involving the displacement, velocity, andacceleration along the path may be obtained by eliminating the time differential dt between Eqs. 12–1 and 12–2. Realize that although we canthen establish another equation, by doing so it will not be independent ofEqs. 12–1 and 12–2. Show that

(12–3)a ds = v dv1:+ 2

ft>s2.m>s2¢v = v - v = 0.

¢v = v¿ - vv,v¿

a =d2s

dt21:+ 2

a =dv

dt1:+ 2

a = lim¢t:0

1¢v>¢t2¢v,¢t

¢v = v¿ - v,¢t,¢v

aavg =¢v¢t

¢tP¿,

Fig. 12–1


sP

Acceleration!

(e)

O

P'

a

v v'

sP

Deceleration!

(f)

O

P'

v v'

– a

Fig. 12–1F

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Constant Acceleration, When the acceleration is constant,eachof the three kinematic equations and may be integrated to obtain formulas that relate v, s, and t.

Velocity as a Function of Time. Integrate assuming thatinitially when

(12–4)

Position as a Function of Time. Integrate assuming that initially when

(12–5)

Velocity as a Function of Position. Either solve for t in Eq. 12–4 andsubstitute into Eq. 12–5, or integrate assuming that initially

at

(12–6)

This equation is not independent of Eqs. 12–4 and 12–5 since it can beobtained by eliminating t between these equations.

The magnitudes and signs of and used in the above threeequations are determined from the chosen origin and positive directionof the s axis as indicated by the arrow written at the left of each equation.Also, it is important to remember that these equations are useful onlywhen the acceleration is constant and when Acommon example of constant accelerated motion occurs when a bodyfalls freely toward the earth. If air resistance is neglected and the distanceof fall is short, then the downward acceleration of the body when it isclose to the earth is constant and approximately or The proof of this is given in Example 13.2.

32.2 ft>s2.9.81 m>s2

v = v0.s = s0,t = 0,

ac,v0,s0,

v2 = v02 + 2ac1s - s02

Constant Acceleration1:+ 2

Lv

v0

v dv = Ls

s0

ac ds

s = s0.v = v0

v dv = ac ds,

s = s0 + v0t + 12 act

2

Constant Acceleration1:+ 2

Ls

s0

ds = Lt

01v0 + act2 dt

t = 0.s = s0

v = ds>dt = v0 + act,

v = v0 + act

Constant Acceleration1:+ 2 L

v

v0

dv = Lt

0ac dt

t = 0.v = v0

ac = dv>dt,

ac,ac ds = v dvv = ds>dt,ac = dv>dt,

a = ac .


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IMPORTANT POINTS• Dynamics is concerned with bodies that have accelerated motion.• Kinematics is a study of the geometry of the motion.• Kinetics is a study of the forces that cause the motion.• Rectilinear kinematics refers to straight-line motion.• Speed refers to the magnitude of velocity.• Average speed is the total distance traveled divided by the total

time. This is different from the average velocity which is thedisplacement divided by the time.

• The acceleration, is negative when the particle isslowing down or decelerating.

• A particle can have an acceleration and yet have zero velocity.• The relationship is derived from and

by eliminating dt.

PROCEDURE FOR ANALYSISThe equations of rectilinear kinematics should be applied using thefollowing procedure.

Coordinate System

• Establish a position coordinate s along the path and specify its fixedorigin and positive direction.

• Since motion is along a straight line, the particle’s position,velocity, and acceleration can be represented as algebraic scalars.For analytical work the sense of s, and a is then determinedfrom their algebraic signs.

• The positive sense for each scalar can be indicated by an arrowshown alongside each kinematic equation as it is applied.

Kinematic Equations

• If a relationship is known between any two of the four variables a,s and t, then a third variable can be obtained by using one of the

kinematic equations, or whichrelates all three variables.*

• Whenever integration is performed, it is important that theposition and velocity be known at a given instant in order toevaluate either the constant of integration if an indefinite integralis used, or the limits of integration if a definite integral is used.

• Remember that Eqs. 12–4 through 12–6 have only a limited use.Never apply these equations unless it is absolutely certain that theacceleration is constant.

*Some standard differentiation and integration formulas are given in Appendix A.

a ds = v dv,v = ds>dta = dv>dt,v,

v,

v = ds>dt,a = dv>dta ds = v dv

a = dv>dt,

s

During the time this rocket undergoesrectilinear motion, its altitude as a functionof time can be measured and expressed as

Its velocity can then be foundusing and its acceleration canbe determined from a = dv>dt.

v = ds>dt,s = s1t2.

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s

O

a, v

Fig. 12–2

E X A M P L E 12.1

The car in Fig. 12–2 moves in a straight line such that for a shorttime its velocity is defined by where t is inseconds. Determine its position and acceleration when When s = 0.t = 0,

t = 3 s.v = 13t2 + 2t2 ft>s,

Solution

Coordinate System. The position coordinate extends from the fixedorigin O to the car, positive to the right.

Position. Since the car’s position can be determined fromsince this equation relates v, s, and t. Noting that

when we have*

When

Ans.

Acceleration. Knowing the acceleration is determinedfrom since this equation relates a, v, and t.

When

Ans.

The formulas for constant acceleration cannot be used to solve thisproblem. Why?

*The same result can be obtained by evaluating a constant of integration C rather thanusing definite limits on the integral. For example, integrating yields

Using the condition that at then C = 0.s = 0,t = 0,s = t3 + t2 + C.ds = 13t2 + 2t2 dt

a = 6132 + 2 = 20 ft>s2 :t = 3 s,

= 6t + 2

a =dv

dt=

d

dt13t2 + 2t21+:2 a = dv>dt,

v = f1t2,s = 1323 + 1322 = 36 ft

t = 3 s,

s = t3 + t2

s `0

s

= t3 + t2 `0

t

Ls

0ds = L

t

013t2 + 2t2 dt

v =ds

dt= 13t2 + 2t21+:2 t = 0,

s = 0v = ds>dt,v = f1t2,

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A small projectile is fired vertically downward into a fluid medium withan initial velocity of 60 m s.Due to the resistance of the fluid the projectileexperiences a deceleration equal to where v is in m s. *Determine the projectile’s velocity and position 4 s after it is fired.

Solution

Coordinate System. Since the motion is downward, the positioncoordinate is positive downward, with origin located at O, Fig. 12–3.

Velocity. Here and so we must determine the velocity asa function of time using since this equation relates v, a,and t. (Why not use ) Separating the variables andintegrating, with when yields

Here the positive root is taken, since the projectile is moving downward.When

Ans.

Position. Knowing we can obtain the projectile’s positionfrom since this equation relates s, v, and t. Using the initialcondition when we have

When Ans.

*Note that to be dimensionally homogeneous, the constant 0.4 has units of s>m2.s = 4.43 m

t = 4 s,

s =1

0.4e c 116022 + 0.8t d1>2 -

160f m

s =2

0.8c 116022 + 0.8t d1>2 `

0

tL

s

0ds = L

t

0c 116022 + 0.8t d-1>2

dt

v =ds

dt= c 116022 + 0.8t d-1>21+p2

t = 0,s = 0,v = ds>dt,

v = f1t2,v = 0.559 m>spt = 4 s,

v = e c 116022 + 0.8t d-1>2 f m>s1

0.8c 1

v2 -116022 d = t

1-0.4

a 1-2b 1

v2 `60

v

= t - 0

Lv

60 m>s dv

-0.4v3 = Lt

0dt

a =dv

dt= -0.4v31+p2 t = 0,v0 = 60 m>sv = v0 + act?

a = dv>dt,a = f1v2

> a = 1-0.4v32m>s2,>

s

O

Fig. 12–3

E X A M P L E 12.2

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During a test a rocket is traveling upward at 75 m s, and when it is 40 mfrom the ground its engine fails. Determine the maximum height reached by the rocket and its speed just before it hits the ground.Whilein motion the rocket is subjected to a constant downward accelerationof due to gravity. Neglect the effect of air resistance.

Solution

Coordinate System. The origin O for the position coordinate s istaken at ground level with positive upward, Fig. 12–4.

Maximum Height. Since the rocket is traveling upward,when At the maximum height the

velocity For the entire motion, the acceleration is(negative since it acts in the opposite sense to positive

velocity or positive displacement). Since is constant the rocket’sposition may be related to its velocity at the two points A and B onthe path by using Eq. 12–6, namely,

Ans.

Velocity. To obtain the velocity of the rocket just before it hits theground, we can apply Eq. 12–6 between points B and C, Fig. 12–4.

Ans.

The negative root was chosen since the rocket is moving downward.Similarly, Eq. 12–6 may also be applied between points A and C,

i.e.,

Note: It should be realized that the rocket is subjected to adeceleration from A to B of and then from B to C it isaccelerated at this rate. Furthermore, even though the rocketmomentarily comes to rest at B the acceleration at B is

downward!9.81 m>s21vB = 029.81 m>s2,

vC = -80.1 m>s = 80.1 m>sp= 175 m>s22 + 21-9.81 m>s2210 - 40 m2vC

2 = vA2 + 2ac1sC - sA21+q2

vC = -80.1 m>s = 80.1 m>sp= 0 + 21-9.81 m>s2210 - 327 m2vC2 = vB

2 + 2ac1sC - sB21+q2sB = 327 m

0 = 175 m>s22 + 21-9.81 m>s221sB - 40 m2vB2 = vA

2 + 2ac1sB - sA21+q2ac

ac = -9.81 m>s2vB = 0.

s = sBt = 0.vA = +75 m>s

9.81 m>s2

sB

>E X A M P L E 12.3

A

O

vB = 0

vA = 75 m/s

sA = 40 m

s

sB

B

C

Fig. 12–4

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A metallic particle is subjected to the influence of a magnetic field asit travels downward through a fluid that extends from plate A to plateB, Fig. 12–5. If the particle is released from rest at the midpoint C,

and the acceleration is where s is inmeters, determine the velocity of the particle when it reaches plate B,

and the time it needs to travel from C to B.

Solution

Coordinate System. As shown in Fig. 12–5, s is taken positivedownward, measured from plate A.

Velocity. Since the velocity as a function of position can beobtained by using Why not use the formulas for constantacceleration? Realizing that at we have

(1)

At

Ans.

The positive root is chosen since the particle is traveling downward,i.e., in the direction.

Time. The time for the particle to travel from C to B can be obtainedusing and Eq. 1, where when FromAppendix A,

At

Ans.t =ln1410.222 - 0.01 + 0.22 + 2.33

2= 0.658 s

s = 200 mm = 0.2 m,

ln14s2 - 0.01 + s2 + 2.33 = 2t

ln14s2 - 0.01 + s2 `0.1

s

= 2t `0

t

Ls

0.1

ds1s2 - 0.0121>2 = Lt

02 dt

= 21s2 - 0.0121>2 dt

ds = v dt1+p2t = 0.s = 0.1 mv = ds>dt

+s

vB = 0.346 m>s = 346 mm>sps = 200 mm = 0.2 m,

v = 21s2 - 0.0121>212 v2 `

0

v

=42

s2 `0.1

s

Lv

0v dv = L

s

0.14s ds

v dv = a ds1+p2 s = 100 mm = 0.1 m,v = 0v dv = a ds.

a = f1s2,s = 200 mm,

a = 14s2m>s2,s = 100 mm,


E X A M P L E 12.4

A

200 mm

100 mm

B

sC

Fig. 12–5

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A particle moves along a horizontal path with a velocity ofwhere t is the time in seconds. If it is initially located

at the origin O, determine the distance traveled in 3.5 s, and the particle’saverage velocity and average speed during the time interval.

Solution

Coordinate System. Here we will assume positive motion to theright, measured from the origin O, Fig. 12–6a.

Distance Traveled. Since the position as a function of timemay be found by integrating with

(1)

In order to determine the distance traveled in 3.5 s, it is necessaryto investigate the path of motion. The graph of the velocity function,Fig. 12–6b, reveals that for the velocity is negative, whichmeans the particle is traveling to the left, and for the velocityis positive, and hence the particle is traveling to the right. Also,at The particle’s position when and can be determined from Eq. 1. This yields

The path is shown in Fig. 12–6a. Hence, the distance traveled in 3.5 s is

Ans.

Velocity. The displacement from to is

and so the average velocity is

Ans.

The average speed is defined in terms of the distance traveled Thispositive scalar is

Ans.1vsp2avg =sT

¢t=

14.1253.5 - 0

= 4.04 m>ssT.

vavg =¢s

¢t=

6.123.5 - 0

= 1.75 m>s :

¢s = s ƒ t = 3.5 s - s ƒ t = 0 = 6.12 - 0 = 6.12 m

t = 3.5 st = 0

sT = 4.0 + 4.0 + 6.125 = 14.125 m = 14.1 m

s ƒ t = 0 = 0 s ƒ t = 2 s = -4.0 m s ƒ t = 3.5 s = 6.125 m

t = 3.5 st = 2 s,t = 0,t = 2 s.v = 0

t 7 2 s0 … t 6 2 s

s = 1t3 - 3t22mL

s

0ds = 3L

t

0t2 dt - 6L

t

0t dt

= 13t2 - 6t2 dtds = v dt1+:2 s = 0.t = 0,v = ds>dt

v = f1t2,

v = 13t2 - 6t2m>s,

E X A M P L E 12.5

O

s = –4.0 m s = 6.125 m

t = 2 s t = 0 s t = 3.5 s

(a)

Fig. 12–6

(0, 0)

v (m/s)

v = 3t2 – 6t

(2s, 0)t (s)

(1 s, –3 m/s)

(b)

Fig. 12–6

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PROBLEMS • 15

s v

Prob. 12–6

P R O B L E M S

12-1. A bicyclist starts from rest and after travelingalong a straight path a distance of 20 m reaches a speedof 30 km/h. Determine his acceleration if it is constant.Also, how long does it take to reach the speed of 30 km/h?

12-2. A car starts from rest and reaches a speed of 80 ft/safter traveling 500 ft along a straight road. Determine itsconstant acceleration and the time of travel.

12-3. A baseball is thrown downward from a 50-ft towerwith an initial speed of 18 ft/s. Determine the speed atwhich it hits the ground and the time of travel.

*12-4. A particle travels along a straight line such thatin 2 s it moves from an initial position to aposition Then in another 4 s it moves from

to Determine the particle’s averagevelocity and average speed during the 6-s time interval.

12-5. Traveling with an initial speed of 70 km/h, a caraccelerates at along a straight road. How longwill it take to reach a speed of 120 km/h? Also, throughwhat distance does the car travel during this time?

12-6. A freight train travels at where t is the elapsed time in seconds. Determine thedistance traveled in three seconds, and the accelerationat this time.

v = 6011 - e-p2 ft>s,6000 km>h2

sC = +2.5 m.sB

sB = -1.5 m.sA = +0.5 m

*12-8. From approximately what floor of a buildingmust a car be dropped from an at-rest position so thatit reaches a speed of when it hits theground? Each floor is 12 ft higher than the one belowit. (Note: You may want to remember this whentraveling

12-9. A car is to be hoisted by elevator to the fourthfloor of a parking garage, which is 48 ft above the ground.If the elevator can accelerate at decelerate at

and reach a maximum speed of 8 ft/s, determinethe shortest time to make the lift, starting from rest andending at rest.

12-10. A particle travels in a straight line such that for ashort time its motion is described by

where a is in If whendetermine the particle’s acceleration when

12-11. The acceleration of a particle as it moves alonga straight line is given by where t is inseconds. If and when determinethe particle’s velocity and position when Also,determine the total distance the particle travels duringthis time period.

*12-12. When a train is traveling along a straight trackat 2 m/s, it begins to accelerate at where v is in m/s. Determine its velocity v and the position3 s after the acceleration.

a = 160 v-42m>s2,

t = 6 s.t = 0,v = 2 m>ss = 1 m

a = 12t - 12m>s2,

t = 3 s.t = 2 s,v = 6 ft>sft>s2.v = 14>a2 ft>s,

2 s … t … 6 s

0.3 ft>s2,0.6 ft>s2,

55 mi>h.280.7 ft>s 155 mi>h2

sv

Prob. 12–12

12-7. The position of a particle along a straight line isgiven by where t is in seconds.Determine its maximum acceleration and maximumvelocity during the time interval 0 … t … 10 s.

s = 1t3 - 9t2 + 15t2 ft,

12-13. The position of a particle along a straight line isgiven by where t is inseconds. Determine the position of the particle when

and the total distance it travels during the 6-s timeinterval. Hint: Plot the path to determine the totaldistance traveled.

t = 6 s

s = 11.5t3 - 13.5t2 + 22.5t2 ft,

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12-14. The position of a particle on a straight line is givenby where t is in seconds.Determinethe position of the particle when and the totaldistance it travels during the 6-s time interval. Hint: Plot thepath to determine the total distance traveled.

12-15. A particle travels to the right along a straight linewith a velocity where s is in meters.Determine its position when if when

*12-16. A particle travels to the right along a straightline with a velocity where s is inmeters. Determine its deceleration when

12-17. Two particles A and B start from rest at the originand move along a straight line such that

and where t isin seconds. Determine the distance between them when

and the total distance each has traveled in

12-18. A car starts from rest and moves along a straightline with an acceleration of where s isin meters. Determine the car’s acceleration when

12-19. A stone A is dropped from rest down a well, andin 1 s another stone B is dropped from rest. Determinethe distance between the stones another second later.

*12-20. A stone A is dropped from rest down a well, andin 1 s another stone B is dropped from rest. Determinethe time interval between the instant A strikes the waterand the instant B strikes the water. Also, at what speeddo they strike the water?

t = 4 s.a = 13s-1>32m>s2,

t = 4 s.t = 4 s

aB = 112t2 - 82 ft>s2,aA = 16t - 32 ft>s2s = 0

s = 2 m.v = [5>14 + s2] m>s,

t = 0.s = 5 mt = 6 sv = [5>14 + s2] m>s,

t = 6 ss = 1t3 - 9t2 + 15t2 ft,

12-21. A particle travels in a straight line withaccelerated motion such that where s is thedistance from the starting point and k is a proportionalityconstant which is to be determined. For thevelocity is 4 ft/s, and for the velocity is 10 ft/s.What is s when

12-22. The acceleration of a rocket traveling upward isgiven by where s is in meters.Determine the rocket’s velocity when and thetime needed to reach this altitude. Initially, and

when t = 0.s = 0v = 0

s = 2 kma = 16 + 0.02s2m>s2,

v = 0?s = 3.5 ft

s = 2 ft

a = -ks,

12-23. The acceleration of a rocket traveling upward isgiven by where s is in meters.Determine the time needed for the rocket to reach analtitude of Initially, and whent = 0.

s = 0v = 0s = 100 m.

a = 16 + 0.02s2m>s2,

80 ft

B

A

Probs. 12–19/20

s

Prob. 12–22

s

Prob. 12–23

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 16

PROBLEMS • 17

40 ft

A

B

5 ft

Prob. 12–26

*12-24. At bullet A is fired vertically with aninitial (muzzle) velocity of 450 m/s. When bulletB is fired upward with a muzzle velocity of 600 m/s.Determine the time t, after A is fired, as to when bulletB passes bullet A. At what altitude does this occur?

■12-25. A particle moves along a straight line with anacceleration of where s is inmeters. Determine the particle’s velocity when if it starts from rest when Use Simpson’s rule toevaluate the integral.

12-26. Ball A is released from rest at a height of 40 ftat the same time that a second ball B is thrown upward5 ft from the ground. If the balls pass one another at aheight of 20 ft, determine the speed at which ball B wasthrown upward.

s = 1 m.s = 2 m,

a = 5>13s1>3 + s5>22m>s2,

t = 3 s,t = 0 *12-28. The acceleration of a particle along a straight

line is defined by where t is in seconds.At and When determine (a) the particle’s position, (b) the total distancetraveled, and (c) the velocity.

12-29. A particle is moving along a straight line such thatwhen it is at the origin it has a velocity of If it beginsto decelerate at the rate of where vis in determine the distance it travels before it stops.

12-30. A particle moves along a straight line with anacceleration of where s is inmeters. Determine the particle’s velocity when if it starts from rest when Use Simpson’s rule toevaluate the integral.

12-31. Determine the time required for a car to travel1 km along a road if the car starts from rest, reaches amaximum speed at some intermediate point, and thenstops at the end of the road. The car can accelerate at

and decelerate at

*12-32. When two cars A and B are next to one another,they are traveling in the same direction with speeds and respectively. If B maintains its constant speed,while A begins to decelerate at determine the distanced between the cars at the instant A stops.

aA,vB,

vA

2 m>s2.1.5 m>s2

s = 1 m.s = 2 m,

a = 5>13s1>3 + s5>22m>s2,

m>s,a = 1-1.5v1>22m>s2,

4 m>s.

t = 9 s,v = 10 m>s.s = 1 mt = 0,a = 12t - 92m>s2,

A B

d

Prob. 12–32

■12-27. A projectile, initially at the origin, movesvertically downward along a straight-line path through afluid medium such that its velocity is defined as

where t is in seconds. Plot theposition s of the projectile during the first 2 s. Use theRunge-Kutta method to evaluate s with incrementalvalues of h = 0.25 s.

v = 318e-t + t21>2 m>s,

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 17


12.3 Rectilinear Kinematics: Erratic Motion

When a particle’s motion during a time period is erratic, it may bedifficult to obtain a continuous mathematical function to describe itsposition, velocity, or acceleration. Instead, the motion may best bedescribed graphically using a series of curves that can be generatedexperimentally from computer output. If the resulting graph describesthe relationship between any two of the variables, a, v, s, t, a graphdescribing the relationship between the other variables can beestablished by using the kinematic equations

Several situations occur frequently.a ds = v dv.v = ds/dt,a = dv>dt,

12-33. If the effects of atmospheric resistance areaccounted for, a freely falling body has an accelerationdefined by the equation where v is in m/s and the positive direction is downward.If the body is released from rest at a very high altitude,determine (a) the velocity when and (b) the body’sterminal or maximum attainable velocity (as ).

12-34. As a body is projected to a high altitude above theearth’s surface, the variation of the acceleration of gravitywith respect to altitude y must be taken into account.Neglecting air resistance, this acceleration is determinedfrom the formula where is theconstant gravitational acceleration at sea level, R is theradius of the earth, and the positive direction is measuredupward. If and determinethe minimum initial velocity (escape velocity) at which aprojectile should be shot vertically from the earth’s surfaceso that it does not fall back to the earth. Hint: This requiresthat as y : q .v = 0

R = 6356 km,g0 = 9.81 m>s2

g0a = -g0[R2>1R + y22],

t : qt = 5 s,

a = 9.81[1 - v2110-42] m>s2,

12-35. Accounting for the variation of gravitationalacceleration a with respect to altitude y (see Prob. 12–34),derive an equation that relates the velocity of a freelyfalling particle to its altitude. Assume that the particle isreleased from rest at an altitude from the earth’ssurface. With what velocity does the particle strike theearth if it is released from rest at an altitude Use the numerical data in Prob. 12–34.

*12-36. When a particle falls through the air, its initialacceleration diminishes until it is zero, andthereafter it falls at a constant or terminal velocity Ifthis variation of the acceleration can be expressed as

determine the time needed forthe velocity to become Initially the particle fallsfrom rest.

v 6 vf.a = 1g>v2

f21v2f - v22, vf.

a = g

y0 = 500 km?

y0

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 18

Given the s–t Graph, Construct the vv–t Graph. If the position of aparticle can be determined experimentally during a time period t, the s–tgraph for the particle can be plotted, Fig. 12–7a.To determine the particle’svelocity as a function of time, i.e., the v–t graph, we must use since this equation relates v, s, and t.Therefore, the velocity at any instantis determined by measuring the slope of the s–t graph, i.e.,

For example, measurement of the slopes at the intermediatepoints (0, 0), on the s–t graph, Fig. 12–7a, gives thecorresponding points on the v–t graph shown in Fig. 12–7b.

It may also be possible to establish the v–t graph mathematically, providedthe segments of the s–t graph can be expressed in the form of equations

Corresponding equations describing the segments of the v–t graphare then determined by time differentiation, since

Given the vv–t Graph, Construct the a–t Graph. When the particle’sv–t graph is known, as in Fig. 12–8a, the acceleration as a function of time,i.e., the a–t graph, can be determined using (Why?) Hence, theacceleration at any instant is determined by measuring the slope of the v–tgraph, i.e.,

For example, measurement of the slopes at the intermediatepoints on the v–t graph, Fig. 12–8a, yieldsthe corresponding points on the a–t graph shown in Fig. 12–8b.

Any segments of the a–t graph can also be determined mathematically,provided the equations of the corresponding segments of the v–t graphare known, This is done by simply taking the time derivativeof since

Since differentiation reduces a polynomial of degree n to that of degreethen if the s–t graph is parabolic (a second-degree curve), the v–t

graph will be a sloping line (a first-degree curve), and the a–t graph willbe a constant or a horizontal line (a zero-degree curve).

n - 1,

a = dv>dt.v = g1t2,v = g1t2.1t3, v321t2, v22,1t1, v12,10, v02, a3a2,a1,a0,

v– t graph = accelerationslope of

dv

dt= a

a = dv>dt.

v = ds>dt.s = f1t2.

1t3, s321t2, s22,1t1, s12, v3v2,v1,v0,

s– t graph = velocityslope of

ds

dt= v

v = ds>dt

Fig. 12–8A

Fig. 12–7A

SECTION 12.3 Rectilinear Kinematics: Erratic Motion • 19

a0 = dv—dt t = 0

v

tt1 t2 t3

v1

v2

v3

v0

= 0

a1 = dv—dt t1

a2 = dv—dt t2

a3 = dv—dt t3

O

(a)

t

a

a0 = 0a1 a2

a3t1 t2 t3O

(b)

Fig. 12–8

tO

v0 = ds!—dt t = 0

v1 = ds!—dt t1

v3 = ds!—dt t3

v2 = ds!—dt t2

s1

s2 s3

t1 t2 t3

(a)

s

v

tO

v1

v2

v3t1 t2

t3

v0

(b)

Fig. 12–7

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 19


Fig. 12–9B

A bicycle moves along a straight road such that its position is describedby the graph shown in Fig. 12–9a. Construct the v–t and a–t graphs for0 … t … 30 s.

t (s)

a (ft/s2)

2

30

(c)

10

Fig. 12–9

t (s)

s (ft)

500

100

10 30

(a)

s = t2

s = 20t – 100 Fig. 12–9

E X A M P L E 12.6

t (s)

v (ft/s)

20

10 30

(b)

v = 2t v = 20

Solution

vv–t Graph. Since the v–t graph can be determined bydifferentiating the equations defining the s–t graph, Fig. 12–9a. We have

The results are plotted in Fig. 12–9b.We can also obtain specific valuesof v by measuring the slope of the s–t graph at a given instant. Forexample, at the slope of the s–t graph is determined fromthe straight line from 10 s to 30 s, i.e.,

a–t Graph. Since the a–t graph can be determined bydifferentiating the equations defining the lines of the v–t graph.This yields

The results are plotted in Fig. 12–9c. Show that whenby measuring the slope of the v–t graph.t = 5 s

a = 2 ft>s2

a =dv

dt= 0v = 2010 6 t … 30 s;

a =dv

dt= 2v = 2t0 … t 6 10 s;

a = dv>dt,

v =¢s

¢t=

500 - 10030 - 10

= 20 ft>st = 20 s;

t = 20 s,

v =ds

dt= 20s = 20t - 10010 s 6 t … 30 s;

v =ds

dt= 2ts = t20 … t 6 10 s;

v = ds>dt,

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 20


Given the a–t Graph, Construct the vv–t Graph. If the a–t graph isgiven, Fig. 12–10a, the v–t graph may be constructed using written in integrated form as

Hence, to construct the v–t graph, we begin by first knowing the particle’sinitial velocity and then add to this small increments of area determined from the a–t graph. In this manner, successive points,

etc., for the v–t graph are determined, Fig. 12–10b. Noticesthat an algebraic addition of the area increments is necessary, since areaslying above the t axis correspond to an increase in v (“positive” area),whereas those lying below the axis indicate a decrease in v (“negative”area).

If segments of the a–t graph can be described by a series of equations,then each of these equations may be integrated to yield equationsdescribing the corresponding segments of the v–t graph. Hence, if thea–t graph is linear (a first-degree curve), integration will yield a v–t graphthat is parabolic (a second-degree curve), etc.

Given the vv–t Graph, Construct the s–t Graph. When the v–t graphis given, Fig. 12–11a, it is possible to determine the s–t graph using

written in integrated form

In the same manner as stated above, we begin by knowing the particle’sinitial position and add (algebraically) to this small area increments

determined from the v–t graph, Fig. 12–11b.If it is possible to describe segments of the v–t graph by a series of

equations, then each of these equations may be integrated to yieldequations that describe corresponding segments of the s–t graph.

¢ss0

displacement = area underv– t graph

¢s = Lv dt

v = ds>dt,

v1 = v0 + ¢v,

1¢v2v0

change in area undervelocity = a–t graph

¢v = La dt

a = dv>dt,

t

a

a0

t1

∆v = a dt0

t1

t

v

v0

t1

v1∆v

(a)

(b)

∫

Fig. 12–10

t

v

v0

t1

∆s = v dt0

t1

t

s

s0

t1

s1∆s

(b)

∫

(a)

Fig. 12–11

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 21


Fig. 12–12B

Fig. 12–12A

The test car in Fig. 12–12a starts from rest and travels along a straighttrack such that it accelerates at a constant rate for 10 s and thendecelerates at a constant rate. Draw the v–t and s–t graphs and determinethe time needed to stop the car. How far has the car traveled?

Solution

vv–t Graph. Since the v–t graph is determined byintegrating the straight-line segments of the a–t graph. Using the initialcondition when we have

When Using this as the initialcondition for the next time period, we have

When we require This yields, Fig. 12–12b,

Ans.

A more direct solution for is possible by realizing that the areaunder the a–t graph is equal to the change in the car’s velocity. Werequire Fig. 12–12a. Thus

Ans.

s–t Graph. Since integrating the equations of the v–tgraph yields the corresponding equations of the s–t graph. Using theinitial conditions when we have

When Using this initial condition,

When the position is

Ans.

The s–t graph is shown in Fig. 12–12c. Note that a direct solution for s ispossible when since the triangular area under the v–t graphwould yield the displacement from to Hence,

Ans.¢s = 12 160211002 = 3000 m

t¿ = 60 s.t = 0¢s = s - 0t¿ = 60 s,

s = -16022 + 1201602 - 600 = 3000 m

t¿ = 60 s,

s = - t2 + 120t - 600

s - 500 = - t2 + 120t - [-11022 + 1201102]Ls

500ds = L

t

101-2t + 1202 dtv = -2t + 120;10 s … t … 60 s;

s = 511022 = 500 m.t = 10 s,

s = 5t2Ls

0ds = L

t

010t dt,v = 10t;0 … t … 10 s;

t = 0,s = 0

ds = v dt,

t¿ = 60 s

0 = 10 m>s2110 s2 + 1-2 m>s221t¿ - 10 s2 = 0

¢v = 0 = A1 + A2,

t¿t¿ = 60 s

v = 0.t = t¿

v = -2t + 120Lv

100dv = L

t

10-2 dt,a = -2;10 s 6 t … t¿;

v = 101102 = 100 m>s.t = 10 s,

v = 10tLv

0dv = L

t

010 dt,a = 10;0 … t 6 10 s;

t = 0,v = 0

dv = a dt,

t¿

t (s)

s (m)

(c)

10 60

500

3000

s = 5t2

s = –t2 + 120t – 600

Fig. 12–12

E X A M P L E 12.7

t (s)

v (m/s)

(b)

100

10

v = 10t

v = –2t + 120

t' = 60

t (s)

a (m/s2)

(a)

10

–210

A1

A2

t'

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 22

Fig. 12–13A

Fig. 12–14A


Given the a–s Graph, Construct the vv–s Graph. In some cases ana–s graph for the particle can be constructed, so that points on the v–sgraph can be determined by using Integrating this equationbetween the limits at and at we have,

Thus, the initial small segment of area under the a–s graph,shown colored in Fig. 12–13a, equals one-half the difference in thesquares of the speed, Therefore, if the area is determinedand the initial value of at is known, then

Fig. 12–13b. Successive points on the v–s graph canbe constructed in this manner starting from the initial velocity

Another way to construct the v–s graph is to first determine theequations which define the segments of the a–s graph. Then thecorresponding equations defining the segments of the v–s graph can beobtained directly from integration, using

Given the vv–s Graph, Construct the a–s Graph. If the v–s graph isknown, the acceleration a at any position s can be determined using

written as

Thus, at any point (s, v) in Fig. 12–14a, the slope dv ds of the v–s graphis measured. Then since v and dv ds are known, the value of a can becalculated, Fig. 12–14b.

We can also determine the segments describing the a–s graphanalytically, provided the equations of the corresponding segments ofthe v–s graph are known. As above, this requires integration usinga ds = v dv.

> >acceleration = velocity times

slope ofv–s graph

a = vadv

dsb

a ds = v dv,

v dv = a ds.

v0.121s1

s0a ds + v0

221>2, v1 =s0 = 0v0

12 1v1

2 - v022. 1s1

s0a ds,

a–s grapharea under

12 1v1

2 - v022 = L

s1

s0

a ds

s = s1,v = v1s = s0v = v0

v dv = a ds.

a

a0

s1

a ds = (v12 – v0

2)0

s1

(a)

1!—2

s

∫

v

v0

s1

v1

(b)

s

Fig. 12–13v

v0

(a)

s

dvds

v

s

a0

(b)

s

a

s

a=v (dv /ds)

Fig. 12–14

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 23


Fig. 12–15A

The v–s graph describing the motion of a motorcycle is shown in Fig. 12–15a. Construct the a–s graph of the motion and determine thetime needed for the motorcycle to reach the position

Solution

a–s Graph. Since the equations for segments of the v–s graph aregiven, the a–s graph can be determined using

The results are plotted in Fig. 12–15b.

Time. The time can be obtained using the v–s graph and because this equation relates v, s, and t. For the first segment ofmotion, at so

At Therefore,for the second segment of motion,

Therefore, at

Ans.t =40050

+ 4.05 = 12.0 s

s = 400 ft,

t =s

50+ 4.05

t - 8.05 =s

50- 4

Lt

8.05dt = L

s

200

ds

50

dt =dsv

=ds

50v = 50;200 ft 6 s … 400 ft;

t = 5 ln[0.212002 + 10] - 5 ln 10 = 8.05 s.s = 200 ft,

t = 5 ln10.2s + 102 - 5 ln 10L

t

0dt = L

s

0

ds

0.2s + 10

dt =dsv

=ds

0.2s + 10v = 0.2s + 10;0 … s 6 200 ft;

t = 0,s = 0

v = ds>dt,

a = vdv

ds= 1502 d

ds1502 = 0

v = 50;200 ft 6 s … 400 ft;

a = vdv

ds= 10.2s + 102 d

ds10.2s + 102 = 0.04s + 2

v = 0.2s + 100 … s 6 200 ft;

a ds = v dv.

s = 400 ft.

E X A M P L E 12.8

v (ft/s)

s (ft)10

50

200 400

v = 0.2s + 10v = 50

(b)

200 400s (ft)

a (ft/s2)

10

2

a = 0.04s + 2

a = 0

Fig. 12–15

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 24

PROBLEMS • 25

P R O B L E M S

12-37. An airplane starts from rest, travels 5000 ft downa runway, and after uniform acceleration, takes off witha speed of 162 mi h. It then climbs in a straight line witha uniform acceleration of until it reaches a constantspeed of 220 mi h. Draw the s–t, v–t, and a–t graphs thatdescribe the motion.

12-38. The elevator starts from rest at the first floor ofthe building. It can accelerate at and thendecelerate at Determine the shortest time it takesto reach a floor 40 ft above the ground.The elevator startsfrom rest and then stops. Draw the a–t, v–t, and s–t graphsfor the motion.

2 ft>s2.5 ft>s2

> 3 ft>s2>

12-39. A freight train starts from rest and travels with aconstant acceleration of After a time itmaintains a constant speed so that when it hastraveled 2000 ft. Determine the time and draw the v–tgraph for the motion.

*12-40. If the position of a particle is defined bywhere t is in seconds, construct

the s–t, v–t, and a–t graphs for 0 … t … 10 s.s = [2 sin1p>52t + 4] m,

t¿t = 160 s

t¿0.5 ft>s2.

40 ft

Prob. 12–38

12-41. The v–t graph for a particle moving through anelectric field from one plate to another has the shapeshown in the figure. The acceleration and decelerationthat occur are constant and both have a magnitude of

If the plates are spaced 200 mm apart, determinethe maximum velocity and the time for the particleto travel from one plate to the other. Also draw the s–tgraph. When the particle is at

12-42. The v–t graph for a particle moving through anelectric field from one plate to another has the shapeshown in the figure, where and Draw the s–t and a–t graphs for the particle. When

the particle is at s = 0.5 m.t = t¿>2 vmax = 10 m>s.t¿ = 0.2 s

s = 100 mm.t = t¿>2 t¿vmax

4 m>s2.

t'/2 t't

v

smax

vmaxs

Probs. 12–41/42

s (m)

a (m/s2)

2

200 300

Prob. 12–43

12-43. The a–s graph for a jeep traveling along a straightroad is given for the first 300 m of its motion. Constructthe v–s graph. At v = 0.s = 0,

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 25


*12-44. A motorcycle starts from rest at andtravels along a straight road with the speed shown by thev–t graph. Determine the motorcycle’s acceleration andposition when and t = 12 s.t = 8 s

s = 0

12-45. An airplane lands on the straight runway, originallytraveling at 110 ft s when If it is subjected to thedecelerations shown, determine the time needed to stopthe plane and construct the s–t graph for the motion.

t¿s = 0.>

12-46. A race cars starting from rest travels along astraight road and for 10 s has the acceleration shown.Construct the v–t graph that describes the motion andfind the distance traveled in 10 s.

4

5v = 5

v = – t + 15v = 1.25t

10 15t (s)

v (m/s)

Prob. 12–44

t (s)5

a (ft/s2)

–3

15 20 t'

–8

Prob. 12–45

6

t (s)

6

6

a (m/s2)

a = t 2

10

1!—6

Prob. 12–46

12-47. The v–t graph for the motion of a train as it movesfrom station A to station B is shown. Draw the a–t graphand determine the average speed and the distance betweenthe stations.

30

40

90 120

v (ft/s)

t (s)

Prob. 12–47

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 26

PROBLEMS • 27

*12-48. The velocity of a car is plotted as shown.Determine the total distance the car moves until it stops

Construct the a–t graph.1t = 80 s2.

12-49. The v–t graph for the motion of a car as if movesalong a straight road is shown. Draw the a–t graph anddetermine the maximum acceleration during the 30-s timeinterval. The car starts from rest at

12-50. The v–t graph for the motion of a car as it movesalong a straight road is shown. Draw the s–t graph anddetermine the average speed and the distance traveledfor the 30-s time interval.The car starts from rest at s = 0.

s = 0.

t (s)

10

40 80

v (m/s)

Prob. 12–48

10 30t (s)

40

60

v (ft/s)

v = t + 30

v = 0.4t2

Probs. 12–49/50

12-51. A car travels along a straight road with the speedshown by the v–t graph. Determine the total distance thecar travels until it stops when Also plot the s–tand a–t graphs.

t = 48 s.

t (s)

6

30 48

v (m/s)

v = – (t – 48)1—3

v = t1—5

Prob. 12–51

*12-52. A man riding upward in a freight elevatoraccidentally drops a package off the elevator when it is100 ft from the ground. If the elevator maintains aconstant upward speed of 4 ft s, determine how high theelevator is from the ground the instant the package hitsthe ground. Draw the v–t curve for the package duringthe time it is in motion. Assume that the package wasreleased with the same upward speed as the elevator.

12-53. Two cars start from rest side by side and travelalong a straight road. Car A accelerates at for 10 sand then maintains a constant speed. Car B acceleratesat until reaching a constant speed of 25 m s andthen maintains this speed. Construct the a–t, v–t, and s–tgraphs for each car until What is the distancebetween the two cars when t = 15 s?

t = 15 s.

>5 m>s2

4 m>s2

>

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 27


12-54. A two-stage rocket is fired vertically from rest atwith an acceleration as shown. After 30 s the first

stage A burns out and the second stage B ignites. Plot thev–t and s–t graphs which describe the motion of thesecond stage for 0 … t … 60 s.

s = 0

15

a (m/s2)

a = 0.01t2

t (s)30 60

9

A

B

Prob. 12–54

■12-55. The a–s graph for a boat moving along a straightpath is given. If the boat starts at when determine its speed when it is at and 125 ft,respectively. Use Simpson’s rule with to evaluatev at s = 125 ft.

n = 100s = 75 ft,

v = 0,s = 0

*12-56. The jet plane starts from rest at and issubjected to the acceleration shown. Determine the speedof the plane when it has traveled 200 ft. Also, how muchtime is required for it to travel 200 ft?

s = 0

12-57. The v–t graph of a car while traveling along aroad is shown. Draw the s–t and a–t graphs for the motion.

s (ft)

a (ft/s2)

100

5

a = s + 6( s – 10)5/3

Prob. 12–55

75

500

a (ft/s2)

a = 75 – 0.15s

s (ft)

Prob. 12–56

20

20 305t (s)

v (m/s)

Prob. 12–57

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 28

PROBLEMS • 29

12-58. A motorcyclist at A is traveling at whenhe wishes to pass the truck T which is traveling at aconstant speed of To do so the motorcyclistaccelerates at until reaching a maximum speed of

If he then maintains this speed, determine thetime needed for him to reach a point located 100 ft infront of the truck. Draw the v–t and s–t graphs for themotorcycle during this time.

85 ft>s.6 ft>s2

60 ft>s.

60 ft>s

12-59. The v–s graph for a go-cart traveling on a straightroad is shown. Determine the acceleration of the go-cartat and Draw the a–s graph.s = 150 m.s = 50 m

*12-60. The v–s graph for the car is given for the first 500ft of its motion. Construct the a–s graph for How long does it take to travel the 500-ft distance? The carstarts at when t = 0.s = 0

0 … s … 500 ft.

100 ft55 ft40 ft

A

T

(vm)1 = 60 ft /s (vm)2 = 85 ft /s

vt = 60 ft /s

Prob. 12–58

8

100s (m)

v (m/s)

200

Prob. 12–59

s (ft)

v (ft/s)

500

60

10

v = 0.1s + 10

Prob. 12–60

12-61. The a–s graph for a train traveling along a straighttrack is given for the first 400 m of its motion. Plot thev–s graph. at s = 0.v = 0

200 400

2

s (m)

a (m/s2)

Prob. 12–61

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 29


*12-64. The test car starts from rest and is subjected toa constant acceleration of for The brakes are then applied, which causes a decelerationat the rate shown until the car stops. Determine the car’smaximum speed and the time t when it stops.

0 … t 6 10 s.ac = 15 ft>s2

12-65. The a–s graph for a race car moving along astraight track has been experimentally determined. If thecar starts from rest at determine its speed when

150 ft, and 200 ft, respectively.s = 50 ft.s = 0,

a (ft/s2)

t (s)t

15

101

2

Prob. 12–64

s (ft)

a (ft/s2)

150

10!

5

200

Prob. 12–65

12-62. The v–s graph for an airplane traveling on a straightrunway is shown. Determine the acceleration of the planeat and Draw the a–s graph.s = 150 m.s = 100 m

12-63. Starting from rest at a boat travels in astraight line with an acceleration as shown by the a–s graph.Determine the boat’s speed when 90, and 200 ft.s = 40,

s = 0,

50

40

200100s (m)

v (m/s)

v = 0.1s + 30

v = 0.4s

Prob. 12–62

50

4

2

150 250s (ft)

a (ft/s2)

Prob. 12–63

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 30

SECTION 12.4 General Curvilinear Motion • 31

12.4 General Curvilinear Motion

Curvilinear motion occurs when the particle moves along a curved path.Since this path is often described in three dimensions, vector analysiswill be used to formulate the particle’s position, velocity, andacceleration.* In this section the general aspects of curvilinear motionare discussed, and in subsequent sections three types of coordinatesystems often used to analyze this motion will be introduced.

Position. Consider a particle located at point P on a space curve definedby the path function s, Fig. 12–16a.The position of the particle, measuredfrom a fixed point O, will be designated by the position vectorThis vector is a function of time since, in general, both its magnitude anddirection change as the particle moves along the curve.

Displacement. Suppose that during a small time interval the particlemoves a distance along the curve to a new position defined by

Fig. 12–16b. The displacement represents the change inthe particle’s position and is determined by vector subtraction; i.e.,

Velocity. During the time the average velocity of the particle isdefined as

The instantaneous velocity is determined from this equation by lettingand consequently the direction of approaches the tangent to

the curve at point P. Hence, or

(12–7)

Since dr will be tangent to the curve at P, the direction of v is also tangentto the curve, Fig. 12–16c. The magnitude of v, which is called the speed,may be obtained by noting that the magnitude of the displacement is the length of the straight line segment from P to Fig. 12–16b.Realizing that this length, approaches the arc length as we have or

(12–8)

Thus, the speed can be obtained by differentiating the path function swith respect to time.

*A summary of some of the important concepts of vector analysis is given inAppendix C.

v =ds

dt

v = lim¢t:0

1¢r>¢t2 = lim¢t:0

1¢s>¢t2, ¢t : 0,¢s¢r,P¿,

¢r

v =drdt

v = lim¢t:0

1¢r>¢t2¢r¢t : 0,

vavg =¢r¢t

¢t,

¢r = r¿ - r.

¢rr¿ = r + ¢r,P¿,¢s¢t

r = r1t2. s

P

rO

Path

Position!

(a)s

Fig. 12–16A

Displacement

(b)

P

r

P'

r'

∆s∆r

s

O

Fig. 12–16B

Velocity

(c)

P

r

v

s

O

Fig. 12–16

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 31


P

v

P'

v'

(d)

Fig. 12–16D

v

v'

(e)

∆v

O'

Fig. 12–16E

va

(f)

O'

Hodograph

Fig. 12–16F

P

Acceleration!

(g)

a

path

Fig. 12–16

Acceleration. If the particle has a velocity v at time t and a velocityat Fig. 12–16d, then the average acceleration of the

particle during the time interval is

where To study this time rate of change, the two velocityvectors in Fig. 12–16d are plotted in Fig. 12–16e such that their tails arelocated at the fixed point and their arrowheads touch points on thecurve. This curve is called a hodograph, and when constructed, itdescribes the locus of points for the arrowhead of the velocity vector inthe same manner as the path s describes the locus of points for thearrowhead of the position vector, Fig. 12–16a.

To obtain the instantaneous acceleration, let in the aboveequation. In the limit will approach the tangent to the hodograph, andso or

(12–9)

Substituting Eq. 12–7 into this result, we can also write

By definition of the derivative, a acts tangent to the hodograph, Fig.12–16f, and therefore, in general, a is not tangent to the path of motion,Fig. 12–16g.To clarify this point, realize that and consequently a mustaccount for the change made in both the magnitude and direction of thevelocity v as the particle moves from P to Fig. 12–16d. Just amagnitude change increases (or decreases) the “length” of v, and this initself would allow a to remain tangent to the path. However, in order forthe particle to follow the path, the directional change always “swings”the velocity vector toward the “inside” or “concave side” of the path,and therefore a cannot remain tangent to the path. In summary, v isalways tangent to the path and a is always tangent to the hodograph.

P¿,

¢v

a =d2rdt2

a =dvdt

a = lim¢t:0

1¢v>¢t2,¢v¢t : 0

O¿

¢v = v¿ - v.

aavg =¢v¢t

¢tt + ¢t,v¿ = v + ¢v

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 32

SECTION 12.5 Curvilinear Motion: Rectangular Components • 33

12.5 Curvilinear Motion: Rectangular Components

Occasionally the motion of a particle can best be described along a paththat is represented using a fixed x, y, z frame of reference.

Position. If at a given instant the particle P is at point (x, y, z) on thecurved path s, Fig. 12–17a, its location is then defined by the position vector

(12–10)

Because of the particle motion and the shape of the path, the x, y, zcomponents of r are generally all functions of time; i.e.,

so that In accordance with the discussion in Appendix C, the magnitude of r

is always positive and defined from Eq. C–3 as

The direction of r is specified by the components of the unit vector

Velocity. The first time derivative of r yields the velocity v of the particle.Hence,

When taking this derivative, it is necessary to account for changes in boththe magnitude and direction of each of the vector’s components. Thederivative of the i component of v is therefore

The second term on the right side is zero, since the x, y, z reference frameis fixed, and therefore the direction (and the magnitude) of i does notchange with time. Differentiation of the j and k components may becarried out in a similar manner, which yields the final result,

(12–11)

where

(12–12)vx = x#

vy = y#

vz = z#

v =drdt

= vxi + vyj + vzk

d

dt1xi2 =

dx

dti + x

didt

v =drdt

=d

dt1xi2 +

d

dt1yj2 +

d

dt1zk2

ur = r>r.

r = 4x2 + y2 + z2

r = r1t2.z = z1t2,y = y1t2, x = x1t2,r = xi + yj + zk

y

x

z

r = xi + yj + zk

P

z

yx

s

ki

j

Position

(a)

Fig. 12–17A

y

x

z

P

s

Velocity

(b)

v = vxi + vy j + vzk

Fig. 12–17

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 33


The “dot” notation represents the first time derivatives of theparametric equations respectively.

The velocity has a magnitude defined as the positive value of

and a direction that is specified by the components of the unit vectorThis direction is always tangent to the path, as shown in Fig.

12–17b.

Acceleration. The acceleration of the particle is obtained by takingthe first time derivative of Eq. 12–11 (or the second time derivative ofEq. 12–10). Using dots to represent the derivatives of the components,we have

(12–13)

where

(12–14)

Here represent, respectively, the first time derivatives of thefunctions or the second timederivatives of the functions

The acceleration has a magnitude defined by the positive value of

and a direction specified by the components of the unit vector Since a represents the time rate of change in velocity, in general a willnot be tangent to the path, Fig. 12–17c.

ua = a>a.

a = 4ax2 + ay

2 + az2

z = z1t2.y = y1t2,x = x1t2,vz = vz1t2,vy = vy1t2,vx = vx1t2,azay,ax,

az = v#z = z

$ay = v

#y = y

$ax = v

#x = x

$

a =dvdt

= axi + ayj + azk

uv = v>v.

v = 4vx2 + vy

2 + vz2

z = z1t2,y = y1t2,x = x1t2,z#

y#,x

#,

y

x

z

P

s

a = axi + ay j + azk

Acceleration

(c)

Fig. 12–17

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IMPORTANT POINTS• Curvilinear motion can cause changes in both the magnitude and

direction of the position, velocity, and acceleration vectors.

• The velocity vector is always directed tangent to the path.

• In general, the acceleration vector is not tangent to the path, butrather, it is tangent to the hodograph.

• If the motion is described using rectangular coordinates, then thecomponents along each of the axes do not change direction, onlytheir magnitude and sense (algebraic sign) will change.

• By considering the component motions, the direction of motionof the particle is automatically taken into account.

PROCEDURE FOR ANALYSIS

Coordinate System

• A rectangular coordinate system can be used to solve problemsfor which the motion can conveniently be expressed in terms ofits x, y, z components.

Kinematic Quantities

• Since rectilinear motion occurs along each coordinate axis, themotion of each component is found using and

or in cases where the motion is not expressed as afunction of time, the equation can be used.

• Once the x, y, z components of v and a have been determined, themagnitudes of these vectors are found from the Pythagoreantheorem, Eq. C–3, and their directions from the components oftheir unit vectors, Eqs. C–4 and C–5.

a ds = v dva = dv>dt;

v = ds>dt


y

x

As the airplane takes off, its path of motioncan be established by knowing its horizontalposition and its vertical positionor altitude both of which can befound from navigation equipment. Byplotting the results from these equations thepath can be shown, and by taking the timederivatives, the velocity and acceleration ofthe plane at any instant can be determined.

y = y1t2,x = x1t2,

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 35


Fig. 12–18B

Fig. 12–18A

At any instant the horizontal position of the weather balloon in Fig.12–18a is defined by where t is in seconds. If the equationof the path is determine (a) the distance of the balloonfrom the station at A when (b) the magnitude and directionof the velocity when and (c) the magnitude and direction ofthe acceleration when

Solution

Position. When and so

The straight-line distance from A to B is therefore

Ans.

Velocity. Using Eqs. 12–12 and application of the chain rule ofcalculus the components of velocity when are

When the magnitude of velocity is therefore

Ans.

The direction is tangent to the path, Fig. 12–18b, where

Ans.

Acceleration. The components of acceleration are determined fromEqs. 12–14 and application of the chain rule, noting that

We have

Thus

Ans.

The direction of a, as shown in Fig. 12–18c, is

Ans.

Note: It is also possible to obtain and by first expressingand then taking successive time derivatives.y = f1t2 = 18t22>10 = 6.4t2

ayvy

ua = tan-1 12.80

= 90°

a = 41022 + 112.822 = 12.8 ft>s2

= 21822>10 + 21162102>10 = 12.8 ft>s2qay = v

#y =

d

dt12xx

# >102 = 21x# 2x# >10 + 2x1x$2>10

ax = v#x = 0

x$ = d218t2>dt2 = 0.

uv = tan-1vy

vx= tan-1 25.6

8= 72.6°

v = 41822 + 125.622 = 26.8 ft>st = 2 s,

vy = y# =

d

dt1x2>102 = 2xx

# >10 = 21162182>10 = 25.6 ft>sq

vx = x# =

d

dt18t2 = 8 ft>s :

t = 2 s

r = 411622 + 125.622 = 30.2 ft

y = 11622>10 = 25.6 ft

x = 8122 ft = 16 ft,t = 2 s,

t = 2 s.t = 2 s,

t = 2 s,y = x2>10,

x = 18t2 ft,

(c)

a = 12.8 ft/s2

B

θa = 90°

Fig. 12–18

E X A M P L E 12.9

(b)

B

v = 26.8 ft/s

θ v = 72.6°

y

A

B

x

16 ft

(a)

y = x2

—10

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The motion of a box B moving along the spiral conveyor shown in Fig. 12–19 is defined by the position vector

where t is in seconds and the arguments forsine and cosine are in radians Determine the locationof the box when and the magnitudes of its velocity andacceleration at this instant.

Solution

Position. Evaluating r when yields

Ans.

The distance of the box from the origin O is

Ans.

The direction of r is obtained from the components of the unit vector,

Hence, the coordinate direction angles Fig. 12–19, are

Ans.

Ans.

Ans.

Velocity. The velocity is defined by

Hence, when the magnitude of velocity, or the speed, is

Ans.

The velocity is tangent to the path as shown in Fig. 12–19. Its coordinatedirection angles can be determined from

Acceleration. The acceleration a of the box, which is shown in Fig. 12–19, is not tangent to the path. Show that

At Ans.a = 2 m>s2t = 0.75s,

a =dvdt

= 5-2 sin12t2i - 2 cos12t2j6 m>s2

uv = v>v.

= 1.02 m>s= 4[1 cos11.5 rad2]2 + [-1 sin11.5 rad2]2 + 1-0.222v = 4vx2 + vy

2 + vz2

t = 0.75 s

= 51 cos12t2i - 1 sin12t2j - 0.2k6 m>sv =drdt

=d

dt[0.5 sin12t2i + 0.5 cos12t2j - 0.2tk]

g = cos-11-0.2872 = 107°

b = cos-110.06782 = 86.1°

a = cos-110.9552 = 17.2°

g,b,a,

= 0.955i + 0.0678j - 0.287k

ur =rr

=0.4990.522

i +0.03540.522

j -0.1500.522

k

r = 410.49922 + 10.035422 + 1-0.15022 = 0.522 m

= 50.499i + 0.0354j - 0.150k6 m

r ƒ t = 0.75 s = 50.5 sin11.5 rad2i + 0.5 cos11.5 rad2j - 0.210.752k6 m

t = 0.75 s

t = 0.75 s1p rad = 180°2.0.5 cos 12t2j - 0.2tk} m,

r = 50.5 sin12t2i +


E X A M P L E 12.10

y

x

a

O

α

γ

v

z

B

rβ

Fig. 12–19

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12.6 Motion of a Projectile

The free-flight motion of a projectile is often studied in terms of itsrectangular components, since the projectile’s acceleration always acts in thevertical direction. To illustrate the kinematic analysis, consider a projectilelaunched at point as shown in Fig. 12–20.The path is defined in thex–y plane such that the initial velocity is having components and

When air resistance is neglected, the only force acting on the projectileis its weight, which causes the projectile to have a constant downwardacceleration of approximately or *g = 32.2 ft>s2.ac = g = 9.81 m>s2

1v02y.1v02xv0,

1x0, y02,

y

x

a = –gj

(v0)y

(v0)x

v0

vx

vy v

r

y0

y

x0

xFig. 12–20

Horizontal Motion. Since application of the constantacceleration equations, 12–4 to 12–6, yields

The first and last equations indicate that the horizontal component ofvelocity always remains constant during the motion.

Vertical Motion. Since the positive y axis is directed upward, thenApplying Eqs. 12–4 to 12–6, we get

Recall that the last equation can be formulated on the basis of eliminatingthe time t between the first two equations, and therefore only two of theabove three equations are independent of one another.

*This assumes that the earth’s gravitational field does not vary with altitude.

vy2 = 1v02y2 - 2g1y - y021+ q2v2 = v0

2 + 2ac1y - y02; y = y0 + 1v02yt - 12 gt21+ q2y = y0 + v0t + 1

2 act2;

vy = 1v02y - gt1+ q2v = v0 + act;

ay = -g.

vx = 1v02x1:+ 2v2 = v02 + 2ac1s - s02; x = x0 + 1v02xt1:+ 2x = x0 + v0t + 1

2 act2;

vx = 1v02x1:+ 2v = v0 + act;

ax = 0,

Each picture in this sequence is taken after thesame time interval.The red ball falls from rest,whereas the yellow ball is given a horizontalvelocity when released. Notice both balls aresubjected to the same downward accelerationsince they remain at the same elevation at anyinstant.This acceleration causes the differencein elevation to increase between successivephotos. Also, note the horizontal distancebetween successive photos of the yellow ballis constant since the velocity in the horizontaldirection remains constant.

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 38

PROCEDURE FOR ANALYSISFree-flight projectile motion problems can be solved using the followingprocedure.

Coordinate System

• Establish the fixed x, y coordinate axes and sketch the trajectory ofthe particle. Between any two points on the path specify the givenproblem data and the three unknowns. In all cases the accelerationof gravity acts downward. The particle’s initial and final velocitiesshould be represented in terms of their x and y components.

• Remember that positive and negative position, velocity, andacceleration components always act in accordance with theirassociated coordinate directions.

Kinematic Equations

• Depending upon the known data and what is to be determined,a choice should be made as to which three of the following fourequations should be applied between the two points on the pathto obtain the most direct solution to the problem.

Horizontal Motion

• The velocity in the horizontal or x direction is constant, i.e.,and

Vertical Motion

• In the vertical or y direction only two of the following threeequations can be used for solution.

• For example, if the particle’s final velocity is not needed, then thefirst and third of these equations (for y) will not be useful.

vy

vy2 = 1v02y2 + 2ac1y - y02y = y0 + 1v02y t + 1

2 act2

vy = 1v02y + act

x = x0 + 1v02xt

1vx2 = 1v02x,

SECTION 12.6 Motion of a Projectile • 39

Gravel falling off the end of this conveyor beltfollows a path that can be predicted using theequations of constant acceleration. In this waythe location of the accumulated pile can bedetermined. Rectilinear coordinates are usedfor the analysis since the acceleration is onlyin the vertical direction.

To summarize, problems involving the motion of a projectile can haveat most three unknowns since only three independent equations can bewritten; that is, one equation in the horizontal direction and two in thevertical direction. Once and are obtained, the resultant velocity v,which is always tangent to the path, is defined by the vector sum as shownin Fig. 12–20.

vyvx

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A sack slides off the ramp, shown in Fig. 12–21, with a horizontalvelocity of If the height of the ramp is 6 m from the floor,determine the time needed for the sack to strike the floor and therange R where sacks begin to pile up.

12 m>s.

x

y

R

6 m

12 m/sA

BC

a = –g

Fig. 12–21

E X A M P L E 12.11

Solution

Coordinate System. The origin of coordinates is established at thebeginning of the path, point A, Fig. 12–21.The initial velocity of a sackhas components and Also, between pointsA and B the acceleration is Since

the three unknowns are R, and the time of flight Here we do not need to determine

Vertical Motion. The vertical distance from A to B is known, andtherefore we can obtain a direct solution for by using the equation

Ans.

This calculation also indicates that if a sack were released from rest at A,it would take the same amount of time to strike the floor at C, Fig. 12–21.

Horizontal Motion. Since t has been calculated, R is determined asfollows:

Ans.R = 13.3 m

R = 0 + 12 m>s 11.11 s2x = x0 + 1v02xtAB1+:2

tAB = 1.11 s

-6 m = 0 + 0 + 12 1-9.81 m>s22tAB

2

y = y0 + 1v02ytAB + 12 actAB

21+q2 tAB

1vB2y.tAB.1vB2y,12 m>s,

1vB)x = 1vA2x =ay = -9.81 m>s2.1vA2y = 0.1vA2x = 12 m>s

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 40

The chipping machine is designed to eject wood chips at as shown in Fig. 12–22. If the tube is oriented at 30° from the horizontal,determine how high, h, the chips strike the pile if they land on the pile20 ft from the tube.

vO = 25 ft>sSECTION 12.6 Motion of a Projectile • 41

4 ft

O

30°

y

x

20 ft

h

A

vO = 25 ft/s

Fig. 12–22

E X A M P L E 12.12

Solution

Coordinate System. When the motion is analyzed between points Oand A, the three unknowns are represented as the height h, time offlight and vertical component of velocity (Note that

) With the origin of coordinates at O, Fig. 12–22, theinitial velocity of a chip has components of

Also, and Since we donot need to determine we have

Horizontal Motion

Vertical Motion. Relating to the initial and final elevations of achip, we have

Ans.h = 1.81 ft

1h - 4 ft2 = 0 + 112.5 ft>s210.9238 s2 + 12 1-32.2 ft>s2210.9238 s22yA = yO + 1vO2ytOA + 1

2 actOA21+q2

tOA

tOA = 0.9238 s

20 ft = 0 + 121.65 ft>s2tOA

xA = xO + 1vO2xtOA1+:2

1vA2y,ay = -32.2 ft>s2.1vA2x = 1vO2x = 21.65 ft>s1vO2y = 125 sin 30°2 ft>s = 12.5 ft>sq

1vO2x = 125 cos 30°2 ft>s = 21.65 ft>s :

1vA2x = 1vO2x.1vA2y.tOA,

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 41


Fig. 12–23B

The track for this racing event was designed so that riders jump off theslope at 30°, from a height of 1 m. During a race it was observed thatthe rider shown in Fig. 12–23a remained in mid air for 1.5 s. Determinethe speed at which he was traveling off the slope, the horizontal distancehe travels before striking the ground, and the maximum height heattains. Neglect the size of the bike and rider.

E X A M P L E 12.13

Fig. 12–23

30°

A

C

B

y

x

R

h

1 m

(b) (a)

Solution

Coordinate System. As shown in Fig. 12–23b, the origin of thecoordinates is established at A. Between the end points of the pathAB the three unknowns are the initial speed range R, and thevertical component of velocity

Vertical Motion. Since the time of flight and the vertical distancebetween the ends of the path are known, we can determine

Ans.

Horizontal Motion. The range R can now be determined.

Ans.In order to find the maximum height h we will consider the path

AC, Fig. 12–23b. Here the three unknowns become the time of flightthe horizontal distance from A to C, and the height h. At the

maximum height and since is known, we can determineh directly without considering using the following equation.

Ans.Show that the bike will strike the ground at B with a velocity havingcomponents of 1vB2y = 8.02 m>sp1vB2x = 11.6 m>s : ,

h = 3.28 m

1022 = 113.38 sin 30°22 + 21-9.812 [1h - 12 - 0]

1vC2y2 = 1vA2y2 + 2ac[1sC2y - 1sA2y]tAC

vA1vC2y = 0,tAC,

= 17.4 m

R = 0 + 13.38 cos 30°11.521sB2x = 1sA2x + 1vA2xtAB1+:2vA = 13.38 m>s = 13.4 m>s-1 = 0 + vA sin 30° 11.52 + 1

2 1-9.81211.5221sB2y = 1sA2y + 1vA2ytAB + 12 act

2AB1+q2 vA.

vB.vA,

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 42

PROBLEMS • 43

12-66. A particle, originally at rest and located at point(3 ft, 2 ft, 5 ft), is subjected to an acceleration of

Determine the particle’s position(x, y, z) at

12-67. The velocity of a particle is given by where t is in seconds. If

the particle is at the origin when determine themagnitude of the particle’s acceleration when Also, what is the x, y, z coordinate position of the particleat this instant?

■*12-68. A particle is traveling with a velocity ofwhere t is in seconds.

Determine the magnitude of the particle’s displacementfrom to Use Simpson’s rule with to evaluate the integrals. What is the magnitude of theparticle’s acceleration when

12-69. The position of a particle is defined bywhere t is in seconds and

the arguments for the sine and cosine are given in radians.Determine the magnitudes of the velocity andacceleration of the particle when Also, prove thatthe path of the particle is elliptical.

12-70. The car travels from A to B, and then from B to C,as shown in the figure. Determine the magnitude of thedisplacement of the car and the distance traveled.

t = 1 s.

r = 551cos 2t2i + 41sin 2t2j6 m,

t = 2 s?

n = 100t = 3 s.t = 0

v = 531te-0.2ti + 4e-0.8t2j6 m>s,

t = 2 s.t = 0,

516t2i + 4t3j + 15t + 22k6m>s,v =

t = 1 s.a = 56ti + 12t2k6 ft>s2.

2 km

AB

3 km

C

Prob. 12–70

P R O B L E M S

A

C

D

B

x

y

10 m

5 m15 m

Prob. 12–71

Cx

y

B 45°

vA = 20 m/s

vB = 30 m/s

vC = 40 m/s

A

Prob. 12–73

12-71. A particle travels along the curve from A to B in2 s. It takes 4 s for it to go from B to C and then 3 s togo from C to D. Determine its average speed when it goesfrom A to D.

*12-72. A car travels east 2 km for 5 minutes, then north3 km for 8 minutes, and then west 4 km for 10 minutes.Determine the total distance traveled and the magnitudeof displacement of the car. Also, what is the magnitudeof the average velocity and the average speed?

12-73. A car traveling along the straight portions of theroad has the velocities indicated in the figure when itarrives at points A, B, and C. If it takes 3 s to go from Ato B, and then 5 s to go from B to C, determine theaverage acceleration between points A and B andbetween points A and C.

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 43


12-78. The particle travels along the path defined by theparabola If the component of velocity alongthe x axis is where t is in seconds, determinethe particle’s distance from the origin O and themagnitude of its acceleration when When

y = 0.x = 0,t = 0,t = 1 s.

vx = 15t2 ft>s,y = 0.5x2.

12-79. When a rocket reaches an altitude of 40 m itbegins to travel along the parabolic path

where the coordinates are measuredin meters. If the component of velocity in the verticaldirection is constant at determine themagnitudes of the rocket’s velocity and accelerationwhen it reaches an altitude of 80 m.

vy = 180 m>s,

1y - 4022 = 160x,

12-81. The nozzle of a garden hose discharges waterat the rate of 15 m/s. If the nozzle is held at groundlevel and directed from the ground, determinethe maximum height reached by the water and thehorizontal distance from the nozzle to where the waterstrikes the ground.

u = 30°

L L

cc

x

y

v0

y = c sin ( x) ––Lπ

Prob. 12–77

x

y

O

y = 0.5 x2

Prob. 12–78

12-74. A particle moves along the curve suchthat its velocity has a constant magnitude of Determine the x and y components of velocity when theparticle is at

12-75. The path of a particle is defined by andthe component of velocity along the y axis is where both k and c are constants. Determine the x and ycomponents of acceleration.

*12-76. A particle is moving along the curvewhere x and y are in ft. If the velocity

component in the x direction is and remainsconstant, determine the magnitudes of the velocity andacceleration when

12-77. The motorcycle travels with constant speed along the path that, for a short distance, takes the form ofa sine curve. Determine the x and y components of itsvelocity at any instant on the curve.

v0

x = 20 ft.

vx = 2 ft>sy = x - 1x2>4002,vy = ct,

y2 = 4kx,

y = 5 ft.

v = 4 ft>s.y = e2x

40 m

y

x

(y – 40)2 = 160x

Prob. 12–79

*12-80. The girl always throws the toys at an angle of30° from point A as shown. Determine the time betweenthrows so that both toys strike the edges of the pool Band C at the same instant. With what speed must shethrow each toy?

0.25 mB

A

C

4 m

2.5 m

1 m

30°

Prob. 12–80

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 44

PROBLEMS • 45

12-82. The balloon A is ascending at the rateand is being carried horizontally by the wind

at If a ballast bag is dropped from theballoon at the instant determine the time neededfor it to strike the ground.Assume that the bag was releasedfrom the balloon with the same velocity as the balloon.Also,with what speed does the bag strike the ground?

h = 50 m,vw = 20 km>h.

vA = 12 km>h

12-83. Determine the maximum height on the wall towhich the firefighter can project water from the hose, ifthe speed of the water at the nozzle is

■*12-84. Determine the smallest angle measuredabove the horizontal, that the hose should be directed sothat the water stream strikes the bottom of the wall at B.The speed of the water at the nozzle is vC = 48 ft>s.

u,

vC = 48 ft>s.

12-85. From a videotape, it was observed that a profootball player kicked a football 126 ft during a measuredtime of 3.6 seconds. Determine the initial speed of theball and the angle at which it was kicked.u

12-86. During a race the dirt bike was observed to leapup off the small hill at A at an angle of 60° with thehorizontal. If the point of landing is 20 ft away, determinethe approximate speed at which the bike was travelingjust before it left the ground. Neglect the size of the bikefor the calculation.

h

vw = 20 km/h

vA = 12 km/h

A

Prob. 12–82

30 ft

3 ft

h

B

A

θC

vC = 48 ft/s

Probs. 12–83/84

A

v0

θ

126 ft

Prob. 12–85

20 ft

A60°

Prob. 12–86

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 45


12-87. Measurements of a shot recorded on a videotapeduring a basketball game are shown. The ball passedthrough the hoop even though it barely cleared the handsof the player B who attempted to block it. Neglecting thesize of the ball, determine the magnitude of its initialvelocity and the height h of the ball when it passes overplayer B.

vA

*12-88. The snowmobile is traveling at 10 m/s when itleaves the embankment at A. Determine the time of flightfrom A to B and the range R of the trajectory.

12-89. The snowmobile is traveling at 10 m/s when itleaves the embankment at A. Determine the speed atwhich it strikes the ground at B and its maximumacceleration along the trajectory AB.

10 fth

C

B

A

vA30°

5 ft25 ft

7 ft

Prob. 12–87

40°

3

4

5

R

B

A

Prob. 12–88

40°

3

4

5

R

B

A

Prob. 12–89

d

B

A 10°45°

vA = 80 ft/s

Prob. 12–90

12-90. A golf ball is struck with a velocity of 80 ft/s asshown. Determine the distance d to where it will land.

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PROBLEMS • 47

12-91. It is observed that the skier leaves the ramp A atan angle with the horizontal. If he strikes theground at B, determine his initial speed and the timeof flight tAB.

vA

uA = 25°

*12-92. The man stands 60 ft from the wall and throwsa ball at it with a speed Determine the angle

at which he should release the ball so that it strikes thewall at the highest point possible.What is this height? Theroom has a ceiling height of 20 ft.

u

v0 = 50 ft>s.

12-93. The ball at A is kicked with a speed and at an angle Determine the point (x, )where it strikes the ground. Assume the ground has theshape of a parabola as shown.

12-94. The ball at A is kicked such that If itstrikes the ground at B having coordinates

determine the speed at which it is kicked andthe speed at which it strikes the ground.y = -9 ft,

x = 15 ft,uA = 30°.

-yuA = 30°.vA = 80 ft>s

12-95. Determine the horizontal velocity of a tennisball at A so that it just clears the net at B. Also, find thedistance s where the ball strikes the ground.

vA

4 m

vAAθ

100 m

B

A

34

5

Prob. 12–91

60 ft

20 fth

5 ft

v0 = 50 ft/sθ

Prob. 12–92

x

–y

B

A θA

vA

y = –0.04x2

y

x

Probs. 12–93/94

21 fts

7.5 ft

3 ft

vA

C

A

B

Prob. 12–95

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 47

B

θC

θD

5 m

A D

C

Prob. 12–97

80 ft

y

x

1

43

5

2

20 ft

20 ft/s

Prob. 12–98

vA = 40 ft/s

AvA

Aθ B C

15 ft d

Prob. 12–99

12-97. The man at A wishes to throw two darts at thetarget at B so that they arrive at the same time. If eachdart is thrown with a speed of 10 m/s, determine theangles and at which they should be thrown and thetime between each throw. Note that the first dart must bethrown at then the second dart is thrown at uD.uC17uD2,uDuC

12-98. The ball is thrown from the tower with a velocityof 20 ft/s as shown. Determine the x and y coordinatesto where the ball strikes the slope. Also, determine thespeed at which the ball hits the ground.

12-99. The baseball player A hits the baseball atand from the horizontal. When the

ball is directly overhead of player B he begins to rununder it. Determine the constant speed at which B mustrun and the distance d in order to make the catch at thesame elevation at which the ball was hit.

uA = 60°vA = 40 ft>s


*12-96. A boy at O throws a ball in the air with a speedat an angle If he then throws another ball at the

same speed at an angle determine the timebetween the throws so the balls collide in mid air at B.

u2 6 u1,vO

u1.vO

x

y

B

O2θ

1θ

Prob. 12–96

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 48

When the path along which a particle is moving is known, it is oftenconvenient to describe the motion using n and t coordinates which actnormal and tangent to the path, respectively, and at the instantconsidered have their origin located at the particle.

Planar Motion. Consider the particle P shown in Fig. 12–24a, which ismoving in a plane along a fixed curve, such that at a given instant it is atposition s, measured from point O. We will now consider a coordinatesystem that has its origin at a fixed point on the curve, and at the instantconsidered this origin happens to coincide with the location of the particle.The t axis is tangent to the curve at P and is positive in the direction ofincreasing s. We will designate this positive direction with the unit vector

A unique choice for the normal axis can be made by noting thatgeometrically the curve is constructed from a series of differential arcsegments ds, Fig. 12–24b. Each segment ds is formed from the arc of anassociated circle having a radius of curvature (rho) and center ofcurvature The normal axis n is perpendicular to the t axis and isdirected from P toward the center of curvature Fig. 12–24a. Thispositive direction, which is always on the concave side of the curve, will bedesignated by the unit vector The plane which contains the n and taxes is referred to as the osculating plane, and in this case it is fixed in theplane of motion.*

Velocity. Since the particle is moving, s is a function of time. Asindicated in Sec. 12.4, the particle’s velocity v has a direction that is alwaystangent to the path, Fig. 12–24c, and a magnitude that is determined bytaking the time derivative of the path function i.e.,(Eq. 12–8). Hence

(12–15)

where

(12–16)v = s#

v = vut

v = ds>dts = s1t2,

un.

O¿,O¿.

r

ut.

SECTION 12.7 Curvilinear Motion: Normal and Tangential Components • 49

12.7 Curvilinear Motion: Normal and Tangential Components

s

O

O'

n

un

ut

t

P

Position

(a)

Fig. 12–24A

O'

ds

Radius of curvature

(b)

O'

O'ds

ds

ρρ

ρρ

ρρ

Fig. 12–24B

O'

Velocity

(c)

P

ρ ρ

v

Fig. 12–24

*The osculating plane may also be defined as that plane which has the greatest contact withthe curve at a point. It is the limiting position of a plane contacting both the point and the arcsegment ds. As noted above, the osculating plane is always coincident with a plane curve;however, each point on a three-dimensional curve has a unique osculating plane.

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 49


Acceleration. The acceleration of the particle is the time rate of changeof the velocity. Thus,

(12–17)

In order to determine the time derivative note that as the particlemoves along the arc ds in time dt, preserves its magnitude of unity;however, its direction changes, and becomes Fig. 12–24d. As shown inFig. 12–24e, we require Here stretches between thearrowheads of and which lie on an infinitesimal arc of radius Hence, has a magnitude of and its direction is definedby Consequently, and therefore the time derivativebecomes Since Fig. 12–24d, then andtherefore

Substituting into Eq. 12–17, a can be written as the sum of its twocomponents,

(12–18)

where

(12–19)

and

(12–20)

These two mutually perpendicular components are shown in Fig. 12–24f,in which case the magnitude of acceleration is the positive value of

(12–21)a = 4at2 + an

2

an =v2

r

at = v#

or atds = v dv

a = atut + anun

u#

t = u#un =

s#

run =

vr

un

u#

= s# >r,ds = r du,u

#t = u

#un.

dut = duun,un.dut = 112 du,dut

ut = 1.uœt,ut

dutuœt = ut + dut.

uœt,

ut

u#

t,

a = v# = v

#ut + vu

#t

O'

(d)

P

ρρ

ut

u't

un

d θ

ds

Fig. 12–24D

ut

u't

un

dut

d θ

(e)

Fig. 12–24E

an

O'

Acceleration

(f)

Pat

a

Fig. 12–24

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To summarize these concepts, consider the following two special casesof motion.

1. If the particle moves along a straight line, then and fromEq. 12–20, Thus and we can conclude that thetangential component of acceleration represents the time rate ofchange in the magnitude of the velocity.

2. If the particle moves along a curve with a constant speed, thenand Therefore, the normal component

of acceleration represents the time rate of change in the directionof the velocity. Since always acts towards the center ofcurvature, this component is sometimes referred to as thecentripetal acceleration.

As a result of these interpretations, a particle moving along the curvedpath in Fig. 12–25 will have accelerations directed as shown.

an

a = an = v2>r.at = v# = 0

a = at = v#,an = 0.

r: q

O'

un

P

O

ub

ut

b osculating plane

t

n

s

Fig. 12–26

Increasing!speed

at

ana

a

at

an

a = at

Change in !direction of !velocity

Change in !magnitude of !velocity Fig. 12–25

Three-Dimensional Motion. If the particle is moving along a spacecurve, Fig. 12–26, then at a given instant the t axis is uniquely specified;however, an infinite number of straight lines can be constructed normalto the tangent axis at P. As in the case of planar motion, we will choosethe positive n axis directed from P toward the path’s center of curvature

This axis is referred to as the principal normal to the curve at P. Withthe n and t axes so defined, Eqs. 12–15 to 12–21 can be used to determinev and a. Since and are always perpendicular to one another and liein the osculating plane, for spatial motion a third unit vector, definesa binormal axis b which is perpendicular to and Fig. 12–26.

Since the three unit vectors are related to one another by the vectorcross product, e.g., Fig. 12–26, it may be possible to usethis relation to establish the direction of one of the axes, if the directionsof the other two are known. For example, no motion occurs in the direction, and so if this direction and are known, then can bedetermined, where in this case Fig. 12–26. Remember,though, that is always on the concave side of the curve.un

un = ub * ut,unut

ub

ub = ut * un,

un,ut

ub,unut

O¿.

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 51


PROCEDURE FOR ANALYSISCoordinate System• Provided the path of the particle is known, we can establish a set

of n and t coordinates having a fixed origin which is coincident withthe particle at the instant considered.

• The positive tangent axis acts in the direction of motion and thepositive normal axis is directed toward the path’s center of curvature.

• The n and t axes are particularly advantageous for studying the velocityand acceleration of the particle, because the t and n components of aare expressed by Eqs. 12–19 and 12–20, respectively.

Velocity

• The particle’s velocity is always tangent to the path.• The magnitude of velocity is found from the time derivative of the

path function.

Tangential Acceleration• The tangential component of acceleration is the result of the time

rate of change in the magnitude of velocity.This component acts inthe positive s direction if the particle’s speed is increasing or in theopposite direction if the speed is decreasing.

• The relations between v, t and s are the same as for rectilinearmotion, namely,

• If is constant, the above equations,when integrated,yield

Normal Acceleration• The normal component of acceleration is the result of the time rate

of change in the direction of the particle’s velocity.This componentis always directed toward the center of curvature of the path, i.e.,along the positive n axis.

• The magnitude of this component is determined from

• If the path is expressed as the radius of curvature atany point on the path is determined from the equation

The derivation of this result is given in any standard calculus text.

r =[1 + 1dy>dx22]3>2

ƒ d2y>dx2 ƒ

ry = f1x2,an =v2

r

v2 = v02 + 21at2c1s - s02v = v0 + 1at2cts = s0 + v0t + 1

2 1at2ct2

at = 1at2c,at

at = v#

at ds = v dv

at,

v = s#

Motorists traveling along this clover-leafinterchange experience a normal accelera-tion due to the change in direction of theirvelocity. A tangential component of accel-eration occurs when the cars’ speed isincreased or decreased.

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 52

Fig. 12–27A


When the skier reaches point A along the parabolic path in Fig. 12–27a,he has a speed of 6 m/s which is increasing at Determine thedirection of his velocity and the direction and magnitude of hisacceleration at this instant.Neglect the size of the skier in the calculation.

Solution

Coordinate System. Although the path has been expressed in termsof its x and y coordinates, we can still establish the origin of the n, taxes at the fixed point A on the path and determine the componentsof v and a along these axes, Fig. 12–27a.

Velocity. By definition, the velocity is always directed tangent to the path. Since then Hence, at A, v makes an angle of with the x axis, Fig. 12–27.Therefore,

Ans.

Acceleration. The acceleration is determined from However, it is first necessary to determine the radius of

curvature of the path at A (10 m, 5 m). Since then

The acceleration becomes

As shown in Fig. 12–27b,

Thus, so that,

Ans.

Note: By using n, t coordinates, we were able to readily solve thisproblem since the n and t components account for the separatechanges in the magnitude and direction of v.

a = 2.37 m>s212.5°daA

57.5° - 45° = 12.5°

f = tan-1 21.273

= 57.5°

a = 41222 + 11.27322 = 2.37 m>s2

= 52ut + 1.273un6 m>s2

= 2ut +16 m>s2228.28 m

un

aA = v#ut +

v2

run

r =[1 + 1dy>dx22]3>2

ƒ d2y>dx2 ƒ=

[1 + 1 110 x22]3>2ƒ 110 ƒ

`x = 10 m

= 28.28 m

d2y>dx2 = 110,

1v2>r2un.a = .

vut +

45°dvAvA = 6 m>su = tan-1 1 = 45°

dy>dx ƒ x = 10 = 1.dy>dx = 110 x,y = 1

20 x2,

2 m>s2.

E X A M P L E 12.14

5 m

10 m

y

x

vA A

t

n

θ

y = x21!—20

(a)

2 m/s2

1.273 m/s2

45°a

t

n

φ

(b)

Fig. 12–27

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 53


A race car C travels around the horizontal circular track that has aradius of 300 ft, Fig. 12–28. If the car increases its speed at a constantrate of starting from rest, determine the time needed for it toreach an acceleration of What is its speed at this instant?8 ft>s2.

7 ft>s2,

E X A M P L E 12.15

t

n

atan

a

C

r = 300 ft

Fig. 12–28

Solution

Coordinate System. The origin of the n and t axes is coincident withthe car at the instant considered. The t axis is in the direction ofmotion, and the positive n axis is directed toward the center of thecircle. This coordinate system is selected since the path is known.

Acceleration. The magnitude of acceleration can be related to itscomponents using Here Since the velocity as a function of time is

Thus

The time needed for the acceleration to reach is therefore

Solving for the positive value of t yields

Ans.

Velocity. The speed at time is

Ans.v = 7t = 714.872 = 34.1 ft>st = 4.87 s

t = 4.87 s

0.163t2 = 41822 - 17228 = 41722 + 10.163t222a = 4at

2 + an2

8 ft>s2

an =v2

r=17t22300

= 0.163t2 ft>s2

v = 0 + 7t

v = v0 + 1at2ctan = v2>r,at = 7 ft>s2.a = 4at

2 + an2.

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 54

Fig. 12–29B

Fig. 12–29A


The boxes in Fig. 12–29a travel along the industrial conveyor. If a boxas in Fig. 12–29b starts from rest at A and increases its speed suchthat where t is in seconds, determine the magnitudeof its acceleration when it arrives at point B.

Solution

Coordinate System. The position of the box at any instant is definedfrom the fixed point A using the position or path coordinate s, Fig.12–29b. The acceleration is to be determined at B, so the origin of then, t axes is at this point.

Acceleration. To determine the acceleration components andit is first necessary to formulate v and so that they may

be evaluated at B. Since when then

(1)

(2)

The time needed for the box to reach point B can be determined byrealizing that the position of B is Fig.12–29b, and since when we have

Substituting into Eqs. 1 and 2 yields

At B, so that

The magnitude of Fig. 12–29c, is therefore

Ans.aB = 411.13822 + 15.24222 = 5.36 m>s2

aB,

1aB2n =vB

2

rB=13.238 m>s22

2 m= 5.242 m>s2

rB = 2 m,

vB = 0.115.6922 = 3.238 m>s1aB2t = v#B = 0.215.6902 = 1.138 m>s2

tB = 5.690 s

6.142 = 0.0333tB3

L6.142

0ds = L

tB

00.1t2 dt

v =ds

dt= 0.1t2

t = 0sA = 0sB = 3 + 2p122>4 = 6.142 m,

v = 0.1t2

Lv

0dv = L

t

00.2t dt

at = v# = 0.2t

t = 0,vA = 0v#

an = v2>r, at = v#

at = 10.2t2m>s2,

E X A M P L E 12.16

(a)

3 m

s

n2 m

B

t

(b)

A

(c)

B

t

naB

1.138 m/s2

5.242 m/s2

Fig. 12–29

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 55


12-106. The jet plane travels along the vertical parabolicpath. When it is at point A it has a speed of 200 m/s, whichis increasing at the rate of Determine themagnitude of acceleration of the plane when it is at point A.

0.8 m>s2.

12-107. Starting from rest, motorboat travels around thecircular path, at a speed wheret is in seconds. Determine the magnitudes of the boatsvelocity and acceleration when it has traveled 20 m.

v = 10.8t2m>s,r = 50 m,

y = 0.4x2

5 km

10 km

y

x

A

Prob. 12–106

ρ = 50 m

v

Prob. 12–107

P R O B L E M S

*12-100. A car is traveling along a circular curve thathas a radius of 50 m. If its speed is 16 m/s and is increasinguniformly at determine the magnitude of itsacceleration at this instant.

12-101. A car moves along a circular track of radius 250 ft such that its speed for a short period of time

is where t is in seconds.Determine the magnitude of its acceleration when

How far has it traveled in

12-102. At a given instant the jet plane has a speed of400 ft/s and an acceleration of acting in thedirection shown. Determine the rate of increase in theplane’s speed and the radius of curvature of the path.r

70 ft>s2

t = 3 s?t = 3 s.

v = 31t + t22 ft>s,0 … t … 4 s,

8 m>s2,

12-103. A boat is traveling along a circular curve havinga radius of 100 ft. If its speed at is 15 ft/s and isincreasing at determine the magnitude ofits acceleration at the instant

*12-104. A boat is traveling along a circular path havinga radius of 20 m. Determine the magnitude of the boat’sacceleration when the speed is and the rate ofincrease in the speed is

■12-105. Starting from rest, a bicyclist travels around ahorizontal circular path, at a speed of

where t is in seconds. Determinethe magnitudes of his velocity and acceleration when hehas traveled s = 3 m.

v = 10.09t2 + 0.1t2m>s,r = 10 m,

v# = 2 m>s2.

v = 5 m>st = 5 s.

v# = 10.8t2 ft>s2,

t = 0

400 ft/s 60°

a = 70 ft/s2

ρ

Prob. 12–102

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 56

PROBLEMS • 57

*12-108. Starting from rest, the motorboat travelsaround the circular path, at a speed

where t is in seconds. Determine themagnitudes of the boats velocity and acceleration at theinstant t = 3 s.

v = 10.2t22m>s,r = 50 m,

ρ = 50 m

v

Prob. 12–108

12-109. A car moves along a circular track of radius250 ft, and its speed for a short period of time is where t is in seconds. Determine themagnitude of its acceleration when How far hasit traveled in

■*12-110. The car travels along the curved path suchthat its speed is increased by where t isin seconds. Determine the magnitudes of its velocity andacceleration after the car has traveled startingfrom rest. Neglect the size of the car.

s = 18 m

v# = 10.5et2m>s2,

t = 2 s?t = 2 s.

v = 31t + t22 ft>s,0 … t … 2 s

s = 18 m

= 30 mρ

Prob. 12–110

12-111. At a given instant the train engine at E has aspeed of 20 m/s and an acceleration of acting inthe direction shown. Determine the rate of increase in thetrain’s speed and the radius of curvature of the path.r

14 m>s2

*12-112. A toboggan is traveling down along a curvewhich can be approximated by the parabola Determine the magnitude of its acceleration when itreaches point A, where its speed is and it isincreasing at the rate of v

#A = 3 m>s2.

vA = 10 m>s,

y = 0.01x2.

60 m

36 m

y = 0.01x2

y

x

A

Prob. 12–112

v = 20 m/s

a = 14 m/s2 E

75°

ρ

Prob. 12–111

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 57


12-115. The truck travels in a circular path havng a radiusof 50 m at a speed of For a short distance from

its speed is increased by wheres is in meters. Determine its speed and the magnitude ofits acceleration when it has moved s = 10 m.

.v = 10.05s2 m>s2,s = 0,

4 = m>s.

*12-116. The jet plane is traveling with a constant speedof 110 m/s along the curved path. Determine themagnitude of the acceleration of the plane at the instantit reaches point A 1y = 02.

12-117. A train is traveling with a constant speed of 14 m/s along the curved path. Determine the magnitudeof the acceleration of the front of the train, B, at theinstant it reaches point A 1y = 02.

s

240 ft

300 ft

Prob. 12–113

s

240 ft

300 ft

Prob. 12–114

12-113. The automobile is originally at rest at Ifits speed is increased by where t is inseconds, determine the magnitudes of its velocity andacceleration when t = 18 s.

v# = 10.05t22 ft>s2,

s = 0.

12-114. The automobile is originally at rest If itthen starts to increase its speed at where t is in seconds, determine the magnitudes of itsvelocity and acceleration at s = 550 ft.

v# = 10.05t22 ft>s2,

s = 0.

50 m

v = (0.05s) m/s2!

v = 4 m/s

.

Prob. 12–115

y

x

y = 15 ln ( )

A

80 m

x––80

Prob. 12–116

vt = 14 m/s

10 m

x (m)A

B

y (m)

y—15( )

x = 10e

Prob. 12–117

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PROBLEMS • 59

A

B

300 ft 60°

300 ft

Prob. 12–118

B

A

v

θ

5 m

Prob. 12–119

B

A

v

θ

5 m

Prob. 12–120

y

x

2 m

y = 0.4x2

A

Prob. 12–121

12-118. When the motorcyclist is at A, he increases hisspeed along the vertical circular path at the rate of

where t is in seconds. If he starts fromrest at A, determine the magnitudes of his velocity andacceleration when he reaches B.

v# = 10.3t2 ft>s2,

■12-119. The car B turns such that its speed is increasedby where t is in seconds. If the car startsfrom rest when determine the magnitudes of itsvelocity and acceleration when the arm AB rotates

Neglect the size of the car.u = 30°.

u = 0°,v#B = 10.5et2m>s2,

*12-120. The car B turns such that its speed is increasedby where t is in seconds. If the car startsfrom rest when determine the magnitudes of itsvelocity and acceleration when Neglect the sizeof the car. Also, through what angle has it traveled?u

t = 2 s.u = 0°,

v#B = 10.5et2m>s2,

12-121. The box of negligible size is sliding down along acurved path defined by the parabola When it isat the speed is andthe increase in speed is Determine themagnitude of the acceleration of the box at this instant.

dvB>dt = 4 m>s2.vB = 8 m>sA 1xA = 2 m, yA = 1.6 m2, y = 0.4x2.

HIBBMC12_0131416782_2_95_v2 8/19/03 11:36 AM Page 59


12-126. The two particles A and B start at the origin O andtravel in opposite directions along the circular path atconstant speeds and respectively. Determine the time when they collide and themagnitude of the acceleration of B just before this happens.

vB = 1.5 m>s,vA = 0.7 m>s

12-122. The ball is ejected horizontally from the tubewith a speed of 8 m/s. Find the equation of the path,

and then find the ball’s velocity and thenormal and tangential components of acceleration whent = 0.25 s.

y = f1x2,

12-123. The motion of a particle is defined by the equationsm and m, where t is in seconds.

Determine the normal and tangential components of theparticle’s velocity and acceleration when

*12-124. The motorcycle travels along the ellipticaltrack at a constant speed v. Determine the greatestmagnitude of the acceleration if a 7 b.

t = 2 s.

y = 1t22x = 12t + t22

12-125. The two particles A and B start at the origin Oand travel in opposite directions along the circular pathat constant speeds and respectively. Determine in (a) the displacementalong the path of each particle, (b) the position vector toeach particle, and (c) the shortest distance between theparticles.

t = 2 s,vB = 1.5 m>s,vA = 0.7 m>s

b

a

y

x

+ = 1x2

a2y2

b2

Prob. 12–124

y

xO

B

vB = 1.5 m/s

vA = 0.7 m/s

A

5 m

Prob. 12–125

vA = 8 m/s

y

xA

Prob. 12–122

y

xO

B

vB = 1.5 m/s

vA = 0.7 m/s

A

5 m

Prob. 12–126

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PROBLEMS • 61

12-127. The race car has an initial speed atA. If it increases its speed along the circular track at therate where s is in meters, determine thetime needed for the car to travel 20 m. Take r = 150 m.

at = 10.4s2m>s2,

vA = 15 m>s

*12-128. A boy sits on a merry-go-round so that he isalways located at from the center of rotation.Themerry-go-round is originally at rest, and then due torotation the boy’s speed is increased at Determinethe time needed for his acceleration to become

12-129. A particle travels along the path where a, b, c are constants. If the speed of

the particle is constant, determine the x and ycomponents of velocity and the normal component ofacceleration when

■12-130. The ball is kicked with an initial speedat an angle with the horizontal. Find

the equation of the path, and then determinethe ball’s velocity and the normal and tangentialcomponents of its acceleration when t = 0.25 s.

y = f1x2,uA = 40°vA = 8 m>sx = 0.

v = v0,a + bx + cx2,

y =

4 ft>s2.2 ft>s2.

r = 8 ft

■12-131. Particles A and B are traveling counter-clockwise around a circular track at a constant speed of8 m/s. If at the instant shown the speed of A is increasedby where is in meters, determine thedistance measured counterclockwise along the track fromB to A when What is the magnitude of theacceleration of each particle at this instant?

t = 1 s.

sAv#A = 14sA2m>s2,

*12-132. Particles A and B are traveling around acircular track at a speed of 8 m/s at the instant shown. Ifthe speed of B is increased by and at thesame instant A has an increase in speed determine how long it takes for a collision to occur. Whatis the magnitude of the acceleration of each particle justbefore the collision occurs?

v#A = 0.8t m>s2,

v#B = 4 m>s2,

As

ρ

Prob. 12–127

x

y

y

xAA = 40°

vA = 8 m/s

θ

Prob. 12–130

r = 5 m

= 120° sB

sA

A

B

θ

Prob. 12–131

r = 5 m

= 120° sB

sA

A

B

θ

Prob. 12–132

HIBBMC12_0131416782_2_95_v2 8/19/03 11:37 AM Page 61


12.8 Curvilinear Motion: Cylindrical Components

In some engineering problems it is often convenient to express the pathof motion in terms of cylindrical coordinates, r, z. If motion is restrictedto the plane, the polar coordinates r and are used.

Polar Coordinates. We can specify the location of particle P shown in Fig.12–30a using both the radial coordinate r, which extends outward from thefixed origin O to the particle, and a transverse coordinate which is thecounterclockwise angle between a fixed reference line and the r axis. Theangle is generally measured in degrees or radians, where The positive directions of the r and coordinates are defined by the unitvectors and respectively. Here or the radial direction extendsfrom P along increasing r, when is held fixed, and or extends from Pin a direction that occurs when r is held fixed and is increased.Note that thesedirections are perpendicular to one another.

u

+uuuu

+ruruu,ur

u

1 rad = 180°>p.

u,

u

u,

O

P

r

r

ur

θ

uθ

θ

Position

(a)

Fig. 12–30

z

y

r

P

x

Prob. 12–135

50 m

Prob. 12–133

12-133. The truck travels at a speed of 4 m/s along acircular road that has a radius of 50 m. For a short distancefrom its speed is then increased by where s is in meters. Determine its speed and the magnitudeof its acceleration when it has moved s = 10 m.

v# = 10.05s2m>s2,s = 0,

■12-134. A go-cart moves along a circular track ofradius 100 ft such that its speed for a short period of time,

is Determine themagnitude of its acceleration when How far hasit traveled in Use Simpson’s rule with toevaluate the integral.

n = 50t = 2 s?t = 2 s.

v = 6011 - e-t22 ft>s.0 … t … 4 s,

12-135. A particle P travels along an elliptical spiralpath such that its position vector r is defined by

where t isin seconds and the arguments for the sine and cosine aregiven in radians. When determine the coordinatedirection angles and which the binormal axis tothe osculating plane makes with the x, y, and z axes. Hint:Solve for the velocity and acceleration of theparticle in terms of their i, j, k components. The binormalis parallel to Why?vP * aP.

aPvP

g,b,a,t = 8 s,

r = 52 cos10.1t2i + 1.5 sin10.1t2j + 12t2k6m,

HIBBMC12_0131416782_2_95_v2 8/19/03 11:37 AM Page 62

Position. At any instant the position of the particle, Fig. 12–30a, isdefined by the position vector

(12–22)

Velocity. The instantaneous velocity v is obtained by taking the timederivative of r. Using a dot to represent time differentiation, we have

To evaluate notice that changes only its direction with respect totime, since by definition the magnitude of this vector is always one unit.Hence, during the time a change will not cause a change in thedirection of however, a change will cause to become where

Fig. 12–30b. The time change in is then For smallangles this vector has a magnitude and acts in the direction. Therefore, and so

(12–23)

Substituting into the above equation for v, the velocity can be written incomponent form as

(12–24)

where

(12–25)

These components are shown graphically in Fig. 12–30c. The radialcomponent is a measure of the rate of increase or decrease in thelength of the radial coordinate, i.e., whereas the transverse component

can be interpreted as the rate of motion along the circumferenceof a circle having a radius r. In particular, the term is calledthe angular velocity, since it indicates the time rate of change of the angle

Common units used for this measurement are rad/s.Since and are mutually perpendicular, the magnitude of velocity

or speed is simply the positive value of

(12–26)

and the direction of v is, of course, tangent to the path at P, Fig. 12–30c.

v = 41r# 22 + 1ru# 22vuvr

u.

u#

= du>dtvu

r#;

vr

vu = ru#

vr = r#

v = vrur + vuuu

u#

r = u#uu

u#

r = lim¢t:0

¢ur

¢t= a lim

¢t:0

¢u¢tb uu

¢ur = ¢uuu,uu¢ur L 11¢u2¢u

¢ur.uruœr = ur + ¢ur,

uœr,ur¢uur;

¢r¢t,

uru#

r,

v = r# = r

#ur + ru

#r

r = rur

SECTION 12.8 Curvilinear Motion: Cylindrical Components • 63

(b)

∆ur

ur

uθ u'r

∆θ

Fig. 12–30B

O

Pr

vr

vθ

θ

v

Velocity

(c)

Fig. 12–30

O

P

r

r

ur

θ

uθ

θ

Position

(a)

Fig. 12–30A

HIBBMC12_0131416782_2_95_v2 8/19/03 11:37 AM Page 63


Acceleration. Taking the time derivatives of Eq. 12–24, using Eqs.12–25, we obtain the particle’s instantaneous acceleration,

To evaluate the term involving it is necessary only to find the changemade in the direction of since its magnitude is always unity. Duringthe time a change will not change the direction of although achange will cause to become where Fig. 12–30d.The time change in is thus For small angles this vector has amagnitude and acts in the direction; i.e.,Thus,

(12–27)

Substituting this result and Eq. 12–23 into the above equation for a, wecan write the acceleration in component form as

(12–28)

where

(12–29)

The term is called the angular accelerationsince it measures the change made in the angular velocity during aninstant of time. Units for this measurement are

Since and are always perpendicular, the magnitude ofacceleration is simply the positive value of

(12–30)

The direction is determined from the vector addition of its twocomponents. In general, a will not be tangent to the path, Fig. 12–30e.

a = 41r$ - ru#222 + 1ru$ + 2r

#u# 22

auar,rad>s2.

u$

= d2 u>dt2 = d>dt1du>dt2au = ru

$+ 2r

#u#

ar = r$ - ru

#2

a = arur + auuu

u#uu = -u

#ur

.uu = lim

¢t:0

¢uu¢t

= - a lim¢t:0

¢u¢tb ur

¢uu = - ¢uur.-ur,¢uu L 11¢u2 ¢uu.uuuœu = uu + ¢uu,uœ

u,uu¢uuu,¢r¢t,

uuu#u,

a = v# = r

$ur + r

#u#

r + r#u#uu + ru

$uu + ru

#u#u

O

Pr

ar

aθ

θ

a

Acceleration

(e)

Fig. 12–30

(d)

ur

u'θ

uθ

∆uθ

∆θ

Fig. 12–30D

HIBBMC12_0131416782_2_95_v2 8/19/03 11:37 AM Page 64


Cylindrical Coordinates. If the particle P moves along a space curveas shown in Fig. 12–31, then its location may be specified by the threecylindrical coordinates, r z.The z coordinate is identical to that used forrectangular coordinates. Since the unit vector defining its direction, isconstant, the time derivatives of this vector are zero, and therefore theposition, velocity, and acceleration of the particle can be written in termsof its cylindrical coordinates as follows:

Time Derivatives. The equations of kinematics require that we obtain thetime derivatives and in order to evaluate the r and components of v and a. Two types of problems generally occur:

1. If the coordinates are specified as time parametric equations,and then the time derivatives can be found

directly. For example, consider

2. If the time-parametric equations are not given, then it will benecessary to specify the path and find the relationshipbetween the time derivatives using the chain rule of calculus.Consider the following examples.

= 10u#2 + 10uu

$r$ = 10[1u# 2u# + u1u$2]r# = 10uu

#r = 5u2

r = f1u2r$ = 8 u

$= 48t

r# = 8t u

#= 24t2

r = 4t2 u = 18t3 + 62u = u1t2,r = r1t2

uu$

u#,r

$,r

#,

a = 1r$ - ru#22ur + 1ru$ + 2r

#u# 2uu + z

$uz

v = r#ur + ru

#uu + z

#uz

rp = rur + zuz

uz,u,

θ r

z

The spiral motion of this boy can be followedby using cylindrical components. Here theradial coordinate r is constant, the transversecoordinate will increase with time as the boyrotates about the vertical, and his altitude zwill decrease with time.

u

z

r

P

rP

ur

uz

uθ

O

θ

Fig. 12–31

HIBBMC12_0131416782_2_95_v2 8/19/03 11:37 AM Page 65


or

If two of the four time derivatives and are known, then the othertwo can be obtained from the equations for first and second time derivativesof See Example 12.19. In some problems, however, two of thesetime derivatives may not be known; instead the magnitude of the particle’svelocity or acceleration may be specified. If this is the case, Eqs. 12–26 and12–30 and may be usedto obtain the necessary relationships involving and See Example 12.20.


Coordinate System

• Polar coordinates are a suitable choice for solving problems forwhich data regarding the angular motion of the radial coordinater is given to describe the particle’s motion. Also, some paths ofmotion can conveniently be described in terms of these coordinates.

• To use polar coordinates, the origin is established at a fixed point,and the radial line r is directed to the particle.

• The transverse coordinate is measured from a fixed referenceline to the radial line.

Velocity and Acceleration

• Once r and the four time derivatives and have beenevaluated at the instant considered, their values can be substitutedinto Eqs. 12–25 and 12–29 to obtain the radial and transversecomponents of v and a.

• If it is necessary to take the time derivatives of it is veryimportant to use the chain rule of calculus.

• Motion in three dimensions requires a simple extension of theabove procedure to include and

Besides the examples which follow, further examples involving thecalculation of and can be found in the “kinematics” sections ofExamples 13.10 through 13.12.

auar

z$.z

#

r = f1u2,u$

u#,r

$,r

#,

u

u$.u

#,r

$,r

#,

a2 = 1r$ - ru#222 + 1ru$ + 2r

#u# 22][v2 = r

# 2 + 1r# u# 22r = f1u2. u

$u#,r

$,r

#,

r# 2 + rr

$ = 912uu# 2 + u2u$22[1r# 2r# + r1r$2] = 18[12uu# 2u# + u21u$2]2rr

# = 18u2u#

r2 = 6u3

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Fig. 12–32B


The amusement park ride shown in Fig. 12–32a consists of a chair thatis rotating in a horizontal circular path of radius r such that the armOB has an angular velocity and angular acceleration Determinethe radial and transverse components of velocity and acceleration ofthe passenger. Neglect his size in the calculation.

u$.u

#

E X A M P L E 12.17

θ θ, r

O

θ

r

(a)

B

· ··

Fig. 12–32

r

θ

θ··

· ·

a = rθ

v = r

θ ,t

θar = –r 2

n

(b)

Solution

Coordinate System. Since the angular motion of the arm is reported,polar coordinates are chosen for the solution, Fig. 12–32a. Here isnot related to r, since the radius is constant for all

Velocity and Acceleration. Equations 12–25 and 12–29 will be usedfor the solution, and so it is first necessary to specify the first andsecond time derivatives of r and Since r is constant, we have

Thus

Ans.

Ans.

Ans.

Ans.

These results are shown in Fig. 12–32b. Also shown are the n, t axes,which in this special case of circular motion happen to be colinearwith the r and axes, respectively. In particular note that

Also,

au = at =dv

dt=

d

dt1ru# 2 =

dr

dtu#

+ rdu

#

dt= 0 + ru

$

-ar = an =v2

r=1ru# 22

r= ru

#2

v = vu = vt = ru#.u

au = ru$

+ 2r#u#

= ru$ar = r

$ - ru#2 = -ru

#2

vu = ru#vr = r

# = 0

r = r r# = 0 r

$ = 0

u.

u.u

HIBBMC12_0131416782_2_95_v2 8/19/03 11:37 AM Page 67


Fig. 12–33B

Fig. 12–33A

The rod OA in Fig. 12–33a is rotating in the horizontal plane such thatAt the same time, the collar B is sliding outward along

OA so that If in both cases t is in seconds, determinethe velocity and acceleration of the collar when

Solution

Coordinate System. Since time-parametric equations of the path aregiven, it is not necessary to relate r to

Velocity and Acceleration. Determining the time derivatives andevaluating when we have


The magnitude of v is

Ans.

Ans.

As shown in Fig. 12–33c,

The magnitude of a is

Ans.

Ans.f = tan-1a1800700b = 68.7° 1180° - f2 + 57.3° = 169°

a = 4170022 + 1180022 = 1930 mm>s2

= 5-700ur + 1800uu6 mm>s2

= [200 - 1001322]ur + [100162 + 2120023]uu

a = 1r$ - ru#22ur + 1ru$ + 2r

#u# 2uu

d = tan-1a300200b = 56.3° d + 57.3° = 114°

v = 4120022 + 130022 = 361 mm>s= 5200ur + 300uu6mm>s= 200ur + 100132uuv = r

#ur + ru

#uu

r$ = 200 `

t = 1 s= 200 mm>s2 u

$= 6t `

t = 1 s= 6 rad>s2.

r# = 200t `

t = 1 s= 200 mm>s u

#= 3t2 `

t = 1 s= 3 rad>s

r = 100t2 `t = 1 s

= 100 mm u = t3 `t = 1 s

= 1 rad = 57.3°

t = 1 s,

u.

t = 1 s.r = 1100t22mm.

u = 1t32 rad.

E X A M P L E 12.18

(a)

θ

A

Br

O

(b)

r

v = 300 mm/sθ= 57.3°θ

vr = 200 mm/s

v

θ

δ

(c)

r

a = 1800 mm/s2θ

= 57.3°θφ θ

a

ar = 700 mm/s2

Fig. 12–33

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Fig. 12–34C

Fig. 12–34B

Fig. 12–34A

100 m= 4 rad/s·θ

r θ

(a)


The searchlight in Fig. 12–34a casts a spot of light along the face of awall that is located 100 m from the searchlight. Determine themagnitudes of the velocity and acceleration at which the spot appearsto travel across the wall at the instant The searchlight isrotating at a constant rate of

Solution

Coordinate System. Polar coordinates will be used to solve thisproblem since the angular rate of the searchlight is given. To find thenecessary time derivatives it is first necessary to relate r to FromFig. 12–34a, this relation is

Velocity and Acceleration. Using the chain rule of calculus, notingthat and we have

Since then and the above equations,when become


Ans.

As shown in Fig. 12–34c,

Ans.

Note: It is also possible to find a without having to calculate (or ).As shown in Fig. 12–34d, since then by vectorresolution, a = 4525.5>cos 45° = 6400 m>s2.

au = 4525.5 m>s2,arr

$= 6400 m>s2

a = 4ar2 + au

2 = 414525.522 + 14525.522= 54525.5ur + 4525.5uu6 m>s2

= [6788.2 - 141.41422]ur + [141.4102 + 21565.724]uu

a = 1r$ - ru#22ur + 1ru$ + 2r

#u# 2uu

= 800 m>sv = 4vr2 + vu

2 = 41565.722 + 1565.722= 5565.7ur + 565.7uu6 m>s= 565.7ur + 141.4142uuv = r#ur + ru

#uu

r$ = 16001sec 45° tan2 45° + sec3 45°2 = 6788.2

r# = 400 sec 45° tan 45° = 565.7

r = 100 sec 45° = 141.4

u = 45°,u$

= 0,u#

= 4 rad>s = constant,

= 100 sec u tan2 u1u# 22 + 100 sec3 u1u# 22 + 1001sec u tan u2u$+ 100 sec u tan u1u$2r$ = 1001sec u tan u2u# 1tan u2u# + 100 sec u1sec2 u2u# 1u# 2r# = 1001sec u tan u2u# d1tan u2 = sec2 u du,d1sec u2 = sec u tan u du,

r = 100>cos u = 100 sec u

u.

u#

= 4 rad>s.u = 45°.

E X A M P L E 12.19

100 mr θ

θ

θ

θ

v

vr

r

v

(b)

100 mr θ

θa

ar

a

(c)θ

r

θ

θ

θar

(d)

= 45°

a

a = 4525.5 m/s2

Fig. 12–34

HIBBMC12_0131416782_2_95_v2 8/19/03 11:37 AM Page 69


Fig. 12–35A

Due to the rotation of the forked rod, the ball A in Fig. 12–35a travelsaround the slotted path, a portion of which is in the shape of a cardioid,

where is in radians. If the ball’s velocity isand its acceleration is at the instant

determine the angular velocity and angular acceleration of the fork.

Solution

Coordinate System. This path is most unusual, and mathematicallyit is best expressed using polar coordinates, as done here, rather thanrectangular coordinates. Also, and must be determined so r,coordinates are an obvious choice.

Velocity and Acceleration. Determining the time derivatives of rusing the chain rule of calculus yields

Evaluating these results at we have

Since using Eq. 12–26 to determine yields

Ans.

In a similar manner, can be found using Eq. 12–30.

Ans.

Vectors a and v are shown in Fig. 12–35b.

u$

= 18 rad>s2

13022 = 1-2422 + u$

2

30 = 4[-0.51422 - 11422]2 + [1u$

+ 2102142]2

a = 41r$ - ru#222 + 1ru$ + 2r

#u# 22u

$u#

= 4 rad>s4 = 41022 + 11u# 22v = 41r# 22 + 1ru# 22 u#

v = 4 ft>s,

r = 1 ft r# = 0 r

$ = -0.5u#2

u = 180°,

r$ = 0.51cos u2u# 1u# 2 + 0.51sin u2u$r# = 0.51sin u2u#r = 0.511 - cos u2

uu$

u#

u$

u# u = 180°,a = 30 ft>s2v = 4 ft>s ur = 0.511 - cos u2 ft,

E X A M P L E 12.20

A θ

θ· θ··,r

r = 0.5 (1 – cos ) ftθ

(a)

r

θ

a = 30 ft/s2

v = 4 ft/s

(b)

Fig. 12–35

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PROBLEMS • 71

*12-136. The time rate of change of acceleration isreferred to as the jerk, which is often used as a means ofmeasuring passenger discomfort. Calculate this vector,in terms of its cylindrical components, using Eq. 12–32.

12-137. If a particle’s position is described by the polarcoordinates and where t is in seconds and the argument for the sine is inradians, determine the radial and tangential componentsof the particle’s velocity and acceleration when

12-138. A particle is moving along a circular path havinga radius of 4 in. such that its position as a function of timeis given by where is in radians and t is inseconds. Determine the magnitude of the acceleration ofthe particle when

12-139. A car is traveling along the circular curve ofradius At the instant shown, its angular rateof rotation is which is increasing at the rateof Determine the magnitudes of the car’svelocity and acceleration at this instant.u$

= 0.2 rad>s2.u#

= 0.4 rad>s,r = 300 ft.

u = 30°.

uu = cos 2t,

t = 2 s.

u = 12e-t2 rad,r = 411 + sin t2m

a#,

*12-140. If a particle moves along a path such thatand where t is in seconds,

plot the path and determine the particle’s radialand transverse components of velocity and acceleration.

r = f1u2u = 1t>22 rad,r = 12 cos t2 ft

r = 300 ft

A

= 0.2 rad/s2..

= 0.4 rad/s.

θθ

θ

Prob. 12–139

P R O B L E M S

12-141. If a particle’s position is described by thepolar coordinates and where t is in seconds, determine the radial andtangential components of its velocity and accelerationwhen

12-142. A particle is moving along a circular path havinga 400-mm radius. Its position as a function of time is givenby where t is in seconds. Determine themagnitude of the particle’s acceleration when The particle starts from rest when

12-143. A particle moves in the x–y plane such thatits position is defined by where t isin seconds. Determine the radial and tangentialcomponents of the particle’s velocity and accelerationwhen

*12-144. A truck is traveling along the horizontalcircular curve of radius with a constant speed

Determine the angular rate of rotation ofthe radial line r and the magnitude of the truck’sacceleration.

12-145. A truck is traveling along the horizontal circularcurve of radius with a speed of 20 m/s which isincreasing at Determine the truck’s radial andtransverse components of acceleration.

3 m>s2.r = 60 m

u#

v = 20 m>s.r = 60 m

t = 2 s.

r = 52ti + 4t2j6 ft,

u = 0°.u = 30°.

u = 12t22 rad,

t = 1 s.

u = 14t2 rad,r = 12 sin 2u2m

θr = 60 m

θ·

Probs. 12–144/145

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*12-152. At the instant shown, the watersprinkler isrotating with an angular speed and anangular acceleration If the nozzle lies inthe vertical plane and water is flowing through it at aconstant rate of 3 m/s, determine the magnitudes ofthe velocity and acceleration of a water particle as itexits the open end, r = 0.2 m.

u$

= 3 rad>s2.u#

= 2 rad>s

θ

θ· = 6 rad/sr

Prob. 12–150

300 mm

700 mm

θ

400 mm

r

O

z

z

y

x

A

v, a

Prob. 12–151

r = 0.2 m

θ

Prob. 12–152

12-146. A particle is moving along a circular path havinga radius of 6 in. such that its position as a function of timeis given by where is in radians, the argumentfor the sine is in degrees, and t is in seconds. Determinethe acceleration of the particle at The particlestarts from rest at

12-147. The slotted link is pinned at O, and as a resultof the constant angular velocity it drives thepeg P for a short distance along the spiral guide

where is in radians. Determine the radialand transverse components of the velocity andacceleration of P at the instant

*12-148. Solve Prob. 12–147 if the slotted link has anangular acceleration when at

12-149. The slotted link is pinned at O, and as a resultof the constant angular velocity it drives thepeg P for a short distance along the spiral guide

m, where is in radians. Determine thevelocity and acceleration of the particle at the instant itleaves the slot in the link, i.e., when r = 0.5 m.

ur = 10.4 u2 u#

= 3 rad>su = p>3 rad.

u#

= 3 rad>su$

= 8 rad>s2

u = p>3 rad.

ur = 10.4 u2m,

u#

= 3 rad>su = 0°.

u = 30°.

uu = sin 3t,

12-150. A block moves outward along the slot in theplatform with a speed of where t is inseconds.The platform rotates at a constant rate of 6 rad/s.If the block starts from rest at the center, determine themagnitudes of its velocity and acceleration when t = 1 s.

r# = 14t2m>s,

0.5 m

O

r

P

θ

= 3 rad/sr = 0.4θθ·

Probs. 12–147/148/149

12-151. The small washer is sliding down the cord OA.When it is at the midpoint, its speed is 200 mm/s and itsacceleration is Express the velocity andacceleration of the washer at this point in terms of itscylindrical components.

10 mm>s2.

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PROBLEMS • 73

12-153. The automobile is traveling from a parking deckdown along a cylindrical spiral ramp at a constant speedof If the ramp descends a distance of 12 mfor every full revolution; determine themagnitude of the car’s acceleration as it moves along theramp, Hint: For part of the solution, note thatthe tangent to the ramp at any point is at an angle of

from the horizontal.Use this to determine the velocity components and which in turn are used to determine and z

#.u

# vz,vu

f = tan-1112>[2p1102]2 = 10.81°

r = 10 m.

u = 2p rad,v = 1.5 m>s.

10 m

12 m

Prob. 12–153

θ

Aθr = 1 + 0.5 cos

O

r

Prob. 12–154

θ

r

r = 1000!____θ

Prob. 12–155

θ

r

r = 1000!____θ

Prob. 12–156

12-154. Because of telescopic action, the end of theindustrial robotic arm extends along the path of thelimaçon At the instant the arm has an angular rotation which isincreasing at Determine the radial andtransverse components of the velocity and accelerationof the object A held in its grip at this instant.

u$

= 0.25 rad>s2.u#

= 0.6 rad>s,u = p>4,r = 11 + 0.5 cos u2m.

12-155. For a short distance the train travels along atrack having the shape of a spiral, where

is in radians. If it maintains a constant speed determine the radial and transverse components of itsvelocity when u = 19p>42 rad.

v = 20 m>s,u

r = 11000>u2m,

*12-156. For a short distance the train travels along atrack having the shape of a spiral, where

is in radians. If the angular rate is constant,determine the radial and transverse

components of its velocity and acceleration whenu = 19p>42 rad.

u#

= 0.2 rad>s,u

r = 11000>u2m,

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12-162. If the car in Prob. 12–161 is accelerating atat the instant determine the required

angular acceleration of the light at this instant.

12-163. A particle P moves along the spiral pathwhere is in radians. If it maintains a

constant speed of determine the magnitudesand as functions of and evaluate each at u = 1 rad.uvuvr

v = 20 ft>s,ur = 110>u2 ft,

u$r = 3000 ft,15 ft>s2

12-157. The arm of the robot has a fixed length so thatand its grip A moves along the path

where is in radians. If where t is in seconds, determine the magnitudes of thegrip’s velocity and acceleration when

12-158. For a short time the arm of the robot isextending at a constant rate such that when

and where t is inseconds. Determine the magnitudes of the velocity andacceleration of the grip A when t = 3 s.

u = 0.5t rad,z = 14t22 ft,r = 3 ft,r# = 1.5 ft>s

t = 3 s.

u = 10.5t2 rad,uz = 13 sin 4u2 ft,r = 3 ft

12-159. The partial surface of the cam is that of alogarithmic spiral where is inradians. If the cam is rotating at a constant angular rateof determine the magnitudes of the velocityand acceleration of the follower rod at the instant

*12-160. Solve Prob. 12–159, if the cam has an angularacceleration of when its angular velocity is

at u = 30°.u#

= 4 rad>s u$

= 2 rad>s2

u = 30°.

u#

= 4 rad>s,

ur = 140e0.05u2mm,

12-161. The searchlight on the boat anchored 2000 ftfrom shore is turned on the automobile, which is travelingalong the straight road at a constant speed of 80 ft/s.Determine the angular rate of rotation of the light whenthe automobile is from the boat.r = 3000 ft

r

z

θ

A

Probs. 12–157/158

θ

= 4 rad/s·θ

r = 40e0.05 θ

Probs. 12–159/160

2000 ft

θθ

θθ

80 ft/s

r

Prob. 12–161

r

r = θ

P

v

10!––θ

Prob. 12–163

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PROBLEMS • 75

*12-164. A particle travels along the portion of the “four-leaf rose” defined by the equation If theangular velocity of the radial coordinate line is

where t is in seconds, determine the radialand transverse components of the particle’s velocity andacceleration at the instant When u = 0.t = 0,u = 30°.

u#

= 13t22 rad>s,

r = 15 cos 2u2m.

r

θ

r = (5 cos 2 )θ

Prob. 12–164

C

B

A

r

θ

0.2 m 0.2 m

0.6 m

0.2 m

Prob. 12–165

10 m

r = 5 m

Prob. 12–166

θ

B

r

A

20 m20 m

20 m20 m

vB = 30 m/s

Prob. 12–167

■12-165. The double collar C is pin connected such thatone collar slides over a fixed rod and the other slides overa rotating rod AB. If the angular velocity of AB is given as

where t is in seconds, and the pathdefined by the fixed rod is determine the radial and transverse components of thecollar’s velocity and acceleration when When

Use Simpson’s rule to determine at t = 1 s.uu = 0°.t = 0,t = 1 s.

r = ƒ 0.4 sin u + 0.2 ƒ m,u#

= 1e0.5t22 rad>s,

12-166. The roller coaster is traveling down along thespiral ramp with a constant speed If the trackdescends a distance of 10 m for every full revolution,

determine the magnitude of the rollercoaster’s acceleration as it moves along the track,

Hint: For part of the solution, note that thetangent to the ramp at any point is at an angle

from the horizontal. Usethis to determine the velocity components and which in turn are used to determine and z

#.u

# vz,vu

f = tan-1[10>2p152] = 17.66°

r = 5 m.

u = 2p rad,

v = 6 m>s.

12-167. A cameraman standing at A is following themovement of a race car, B, which is traveling around acurved track at a constant speed of 30 m/s. Determine theangular rate at which the man must turn in order to keepthe camera directed on the car at the instant u = 30°.

u#

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*12-168. The pin follows the path described by theequation At the instant

and Determine themagnitudes of the pin’s velocity and acceleration at thisinstant. Neglect the size of the pin.

u$

= 0.5 rad>s2.u#

= 0.7 rad>s u = 30°,r = 10.2 + 0.15 cos u2m.

12-169. For a short time the position of the roller-coastercar along its path is defined by the equations

and where t is inseconds. Determine the magnitude of the car’s velocityand acceleration when t = 4 s.

z = 1-8 cos u2m,u = 10.3t2 rad,r = 25 m,

r

= 0.7 rad/sθ

r = 0.2+ 0.15 cos θ

θx

y

·

Prob. 12–168

r

z

z

θ

Prob. 12–171

A

z

rθ

Prob. 12–169

r

θ

A

G

(vA)x

(vA)y

r = 0.3 + 0.2 cosθ

Prob. 12–170

12-170. The mechanism of a machine is constructed so thatthe roller at A follows the surface of the cam described bythe equation If and

determine the magnitudes of the roller’s velocity andacceleration when Neglect the size of the roller.Also compute the velocity components and ofthe roller at this instant. The rod to which the roller isattached remains vertical and can slide up or down alongthe guides while the guides translate horizontally to the left.

1vA2y1vA2xu = 30°.u$

= 0,u#

= 0.5 rad>sr = 10.3 + 0.2 cos u2m.

12-171. The crate slides down the section of the spiral rampsuch that and where t isin seconds. If the rate of rotation about the z axis is

determine the magnitudes of the velocityand acceleration of the crate at the instant z = 10 ft.

.u = 0.04pt rad>s,

z = 1100 - 0.1t22 ft,r = 10.5z2 ft

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SECTION 12.9 Absolute Dependent Motion Analysis of Two Particles • 77

12.9 Absolute Dependent Motion Analysis of Two Particles

In some types of problems the motion of one particle will depend on thecorresponding motion of another particle. This dependency commonlyoccurs if the particles are interconnected by inextensible cords which arewrapped around pulleys. For example, the movement of block Adownward along the inclined plane in Fig. 12–36 will cause acorresponding movement of block B up the other incline. We can showthis mathematically by first specifying the location of the blocks usingposition coordinates and Note that each of the coordinate axes is(1) referenced from a fixed point (O) or fixed datum line, (2) measuredalong each inclined plane in the direction of motion of block A and blockB, and (3) has a positive sense from C to A and D to B. If the total cordlength is the position coordinates are related by the equation

Here is the length of the cord passing over arc CD.Taking the timederivative of this expression, realizing that and remain constant,while and measure the lengths of the changing segments of thecord, we have

The negative sign indicates that when block A has a velocity downward,i.e., in the direction of positive it causes a corresponding upwardvelocity of block B; i.e., B moves in the negative direction.

In a similar manner, time differentiation of the velocities yields therelation between the accelerations, i.e.,

A more complicated example involving dependent motion of twoblocks is shown in Fig. 12–37a. In this case, the position of block A isspecified by and the position of the end of the cord from which blockB is suspended is defined by Here we have chosen coordinate axeswhich are (1) referenced from fixed points or datums, (2) measured inthe direction of motion of each block, and (3) positive to the right and positive downward During the motion, the red coloredsegments of the cord in Fig. 12–37a remain constant. If l represents thetotal length of cord minus these segments, then the position coordinatescan be related by the equation

Since l and h are constant during the motion, the two time derivatives yield

Hence, when B moves downward A moves to the left withtwo times the motion.

1-sA21+sB2,2vB = -vA 2aB = -aA

2sB + h + sA = l

1sB2. 1sA2sB.sA,

aB = -aA

sB

sA,

dsA

dt+

dsB

dt= 0 or vB = -vA

sBsA

lTlCD

lCD

sA + lCD + sB = lT

lT,

sB.sAA B

C D

DatumDatumsB

sA

O

Fig. 12–36

A

sB

Datum

hB

sADatum

(a)

Fig. 12–37

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This example can also be worked by defining the position of block B fromthe center of the bottom pulley (a fixed point), Fig. 12–37b. In this case

Time differentiation yields

Here the signs are the same. Why?

PROCEDURE FOR ANALYSISThe above method of relating the dependent motion of one particleto that of another can be performed using algebraic scalars orposition coordinates provided each particle moves along a rectilinearpath. When this is the case, only the magnitudes of the velocity andacceleration of the particles will change, not their line of direction.The following procedure is required.

Position-Coordinate Equation

• Establish position coordinates which have their origin located at afixed point or datum.

• The coordinates are directed along the path of motion and extendto a point having the same motion as each of the particles.

• It is not necessary that the origin be the same for each of thecoordinates; however, it is important that each coordinate axisselected be directed along the path of motion of the particle.

• Using geometry or trigonometry, relate the coordinates to thetotal length of the cord, or to that portion of cord, l, whichexcludes the segments that do not change length as the particlesmove—such as arc segments wrapped over pulleys.

• If a problem involves a system of two or more cords wrappedaround pulleys, then the position of a point on one cord must berelated to the position of a point on another cord using the aboveprocedure. Separate equations are written for a fixed length of eachcord of the system and the positions of the two particles are thenrelated by these equations (see Examples 12.22 and 12.23).

Time Derivatives

• Two successive time derivatives of the position-coordinateequations yield the required velocity and acceleration equationswhich relate the motions of the particles.

• The signs of the terms in these equations will be consistent with thosethat specify the positive and negative sense of the position coordinates.

lT,

2vB = vA 2aB = aA

21h - sB2 + h + sA = l

sB

Datum

(b)

A

Datum

hB

sADatum

Fig. 12–37

The motion of the traveling block on this oilrig depends upon the motion of the cableconnected to the winch which operates it. Itis important to be able to relate thesemotions in order to determine the powerrequirements of the winch and the force inthe cable caused by accelerated motion.

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Determine the speed of block A in Fig. 12–38 if block B has an upwardspeed of 6 ft/s.

E X A M P L E 12.21

sA

B

A

sB

C D

E

6 ft/s

Datum

Fig. 12–38

Solution

Position-Coordinate Equation. There is one cord in this systemhaving segments which are changing length. Position coordinates and will be used since each is measured from a fixed point (C orD) and extends along each block’s path of motion. In particular, isdirected to point E since motion of B and E is the same.

The red colored segments of the cord in Fig. 12–38 remain at aconstant length and do not have to be considered as the blocks move.The remaining length of cord, l, is also constant and is related to thechanging position coordinates and by the equation

Time Derivative. Taking the time derivative yields

so that when (upward),

Ans.vA = 18 ft>spvB = -6 ft>s vA + 3vB = 0

sA + 3sB = l

sBsA

sB

sB

sA

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Determine the speed of block A in Fig. 12–39 if block B has an upwardspeed of 6 ft/s.

E X A M P L E 12.22

A

B

sA

sB

Datum

6 ft/s

C

D

sC

Fig. 12–39

Solution

Position-Coordinate Equation. As shown, the positions of blocks Aand B are defined using coordinates and Since the system hastwo cords which change length, it will be necessary to use a thirdcoordinate, in order to relate to In other words, the lengthof one of the cords can be expressed in terms of and and thelength of the other cord can be expressed in terms of and

The red colored segments of the cords in Fig. 12–39 do not have tobe considered in the analysis. Why? For the remaining cord lengths,say and we have

Eliminating yields an equation defining the positions of both blocks,i.e.,

Time Derivative. The time derivative gives

so that when (upward),

Ans.vA = +24 ft>s = 24 ft>spvB = -6 ft>svA + 4vB = 0

sA + 4sB = 2l2 + l1

sC

sA + 2sC = l1 sB + 1sB - sC2 = l2

l2,l1

sC.sB

sC,sA

sB.sAsC,

sB.sA

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Determine the speed with which block B rises in Fig. 12–40 if the endof the cord at A is pulled down with a speed of 2 m/s.

E X A M P L E 12.23

sA

Datum

sC

sB

C

A

B

D

2 m/s

E

Fig. 12–40Solution

Position-Coordinate Equation. The position of point A is definedby and the position of block B is specified by since point E onthe pulley will have the same motion as the block. Both coordinatesare measured from a horizontal datum passing through the fixed pinat pulley D. Since the system consists of two cords, the coordinates and cannot be related directly. Instead, by establishing a thirdposition coordinate, we can now express the length of one of thecords in terms of and and the length of the other cord in termsof and

Excluding the red colored segments of the cords in Fig. 12–40, theremaining constant cord lengths and (along with the hook andlink dimensions) can be expressed as

Eliminating yields

As required, this equation relates the position of block B to theposition of point A.

Time Derivative. The time derivative gives

so that when (downward),

Ans.vB = -0.5 m>s = 0.5 m>sqvA = 2 m>s vA + 4vB = 0

sA

sB

sA + 4sB = l2 + 2l1

sC

1sA - sC2 + 1sB - sC2 + sB = l2

sC + sB = l1

l2l1

sC.sB,sA,sC,sB

sC,sB

sA

sBsA,

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A man at A is hoisting a safe S as shown in Fig. 12–41 by walking tothe right with a constant velocity Determine the velocityand acceleration of the safe when it reaches the elevation at E. Therope is 30 m long and passes over a small pulley at D.

Solution

Position-Coordinate Equation. This problem is unlike the previousexamples since rope segment DA changes both direction and magnitude.However, the ends of the rope, which define the positions of S and A,are specified by means of the x and y coordinates measured from a fixedpoint and directed along the paths of motion of the ends of the rope.

The x and y coordinates may be related since the rope has a fixedlength which at all times is equal to the length of segmentDA plus CD. Using the Pythagorean theorem to determine wehave also, Hence,

(1)

Time Derivatives. Taking the time derivative, using the chain rule,where and yields

(2)

At x is determined from Eq. 1, i.e., Hence, fromEq. 2 with

Ans.

The acceleration is determined by taking the time derivative of Eq. 2. Since is constant, then and we haveaA = dvA>dt = 0,vA

vS =202225 + 12022 10.52 = 0.4 m>s = 400 mm>sqvA = 0.5 m>s,

x = 20 m.y = 10 m,

=x2225 + x2

vA.

vS =dy

dt= B1

22x2225 + x2

R dx

dt

vA = dx>dt,vS = dy>dt

y = 2225 + x2 - 15

30 = 211522 + x2 + 115 - y2l = lDA + lCD

lCD = 15 - y.lDA = 211522 + x2;lDA,

l = 30 m,

vA = 0.5 m>s.

E X A M P L E 12.24

D

10 m

15 m

E

y

vA = 0.5 m/sA

S

C

x

Fig. 12–41

aS =d2y

dt2 = B -x1dx>dt21225 + x223>2RxvA + B 12225 + x2R adx

dtbvA + B 12225 + x2

RxdvA

dt=

225vA21225 + x223>2

At with the acceleration becomes

Ans.

Note that the constant velocity at A causes the other end C of therope to have an acceleration since causes segment DA to changeits direction as well as its length.

vA

aS =22510.5 m>s22

[225 + 120 m22]3>2 = 0.00360 m>s2 = 3.60 mm>s2q

vA = 0.5 m>s,x = 20 m,

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SECTION 12.10 Relative-Motion Analysis of Two Particles Using Translating Axes • 83

12.10 Relative-Motion Analysis of Two Particles Using Translating Axes

Throughout this chapter the absolute motion of a particle has beendetermined using a single fixed reference frame for measurement. Thereare many cases, however, where the path of motion for a particle iscomplicated, so that it may be feasible to analyze the motion in parts byusing two or more frames of reference. For example, the motion of aparticle located at the tip of an airplane propeller, while the plane is inflight, is more easily described if one observes first the motion of theairplane from a fixed reference and then superimposes (vectorially) thecircular motion of the particle measured from a reference attached tothe airplane. Any type of coordinates—rectangular, cylindrical, etc.—may be chosen to describe these two different motions.

In this section only translating frames of reference will be consideredfor the analysis. Relative-motion analysis of particles using rotatingframes of reference will be treated in Secs. 16.8 and 20.4, since such ananalysis depends on prior knowledge of the kinematics of line segments.

Position. Consider particles A and B, which move along the arbitrarypaths aa and bb, respectively, as shown in Fig. 12–42a.The absolute positionof each particle, and is measured from the common origin O of thefixed x, y, z reference frame.The origin of a second frame of reference

is attached to and moves with particle A.The axes of this frame areonly permitted to translate relative to the fixed frame.The relative positionof “B with respect to A” is designated by a relative-position vectorUsing vector addition, the three vectors shown in Fig. 12–42a can berelated by the equation*

(12–33)

Velocity. An equation that relates the velocities of the particles can bedetermined by taking the time derivative of Eq. 12–33, i.e.,

(12–34)

Here and refer to absolute velocities, since theyare observed from the fixed frame; whereas the relative velocity

is observed from the translating frame. It is importantvB>A = drB>A>dt

vA = drA>dtvB = drB>dt

vB = vA + vB>A

rB = rA + rB>A

rB>A.

z¿y¿,x¿,

rB,rA

*An easy way to remember the setup of this equation, and others like it, is to note the“cancellation” of the subscript A between the two terms, i.e., rB = rA + rB>A.

z'

z

A

O

x'

x

y

y'

rB

rA

rB/A

Translating!observer

a

a

b

b

Fixed!observer

(a)

B

Fig. 12–42

HIBBMC12_0131416782_2_95_v2 8/19/03 11:37 AM Page 83


to note that since the axes translate, the components of will not change direction and therefore the time derivative of this vector’scomponents will only have to account for the change in the vector’smagnitude. Equation 12–34 therefore states that the velocity of B is equalto the velocity of A plus (vectorially) the relative velocity of “B withrespect to A,” as measured by the translating observer fixed in the

reference, Fig. 12–42b.

Acceleration. The time derivative of Eq. 12–34 yields a similar vectorrelationship between the absolute and relative accelerations of particlesA and B.

(12–35)

Here is the acceleration of B as seen by the observer located at Aand translating with the reference frame. The vector additionis shown in Fig. 12–42c.


• When applying the relative-position equation, itis first necessary to specify the locations of the fixed x, y, z, andtranslating axes.

• Usually, the origin A of the translating axes is located at a pointhaving a known position, Fig. 12–42a.

• A graphical representation of the vector addition can be shown, and both the known and unknown quantitieslabeled on this sketch.

• Since vector addition forms a triangle, there can be at most twounknowns, represented by the magnitudes and/or directions ofthe vector quantities.

• These unknowns can be solved for either graphically, usingtrigonometry (law of sines, law of cosines), or by resolving eachof the three vectors and into rectangular or Cartesiancomponents, thereby generating a set of scalar equations.

• The relative-motion equations and are applied in the same manner as explained above, except in thiscase the origin O of the fixed x,y,z axes does not have to be specified,Figs. 12–42b and 12–42c.

aB = aA + aB>AvB = vA + vB>ArB>ArA,rB,

rB = rA + rB>ArA,

z¿y¿,x¿,

rB = rA + rB>A,

z¿y¿,x¿,aB>A

aB = aA + aB>A

z¿y¿,x¿,

rB>Az¿y¿,x¿,

vB

vB/A

vA

(b)

Fig. 12–42B

(c)

aB/A

aBaA

Fig. 12–42

The pilots of these jet planes flying close toone another must be aware of their relativepositions and velocities at all times in orderto avoid a collision.

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Fig. 12–43

Fig. 12–43A


A train, traveling at a constant speed of 60 mi/h, crosses over a roadas shown in Fig. 12–43a. If automobile A is traveling at 45 mi/h alongthe road, determine the magnitude and direction of the relativevelocity of the train with respect to the automobile.

Solution I

Vector Analysis. The relative velocity is measured from thetranslating axes attached to the automobile, Fig. 12–43a. It isdetermined from Since and are known in bothmagnitude and direction, the unknowns become the x and ycomponents of Using the x, y axes in Fig. 12–43a and a Cartesianvector analysis, we have

Ans.

The magnitude of is thus

Ans.

From the direction of each component, Fig. 12–43b, the direction ofdefined from the x axis is

Ans.

Note that the vector addition shown in Fig. 12–43b indicates thecorrect sense for This figure anticipates the answer and can beused to check it.

Solution II

Scalar Analysis. The unknown components of can also bedetermined by applying a scalar analysis. We will assume thesecomponents act in the positive x and y directions. Thus,

Resolving each vector into its x and y components yields

Solving, we obtain the previous results,

Ans.1vT>A2y = -31.8 mi>h = 31.8 mi>hp1vT>A2x = 28.2 mi>h = 28.2 mi>h :

0 = 45 sin 45° + 0 + 1vT>A2y1+q2 60 = 45 cos 45° + 1vT>A2x + 01+:2B 60 mi>h

: R = B 45 mi>ha45° R + B 1vT>A2x

: R + B 1vT>A2yq RvT = vA + vT>A

vT>AvT>A.

u = 48.5°cu

tan u =1vT>A2y1vT>A2x =

31.828.2

vT>AvT>A = 2128.222 + 1-31.822 = 42.5 mi>hvT>AvT>A = 528.2i - 31.8j6mi>h60i = 145 cos 45° i + 45 sin 45° j2 + vT>AvT = vA + vT>A

vT>A.

vAvTvT = vA + vT>A.y¿x¿,

vT>A

E X A M P L E 12.25

31.8 mi/h

28.2 mi/h

(b)

θ

vT/A

vT = 60 mi/hy'

45°T

vA = 45 mi/h

y

x

A

(a)

x'

v T = 60 mi/h

(c)

45° θ

vA = 45 mi/h vT/A

Fig. 12–43

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Fig. 12–44B

Fig. 12–44A

(b)

vB/A

vB = 600 km/hvA = 700 km/h

Plane A in Fig. 12–44a is flying along a straight-line path, whereasplane B is flying along a circular path having a radius of curvature of

Determine the velocity and acceleration of B asmeasured by the pilot of A.

Solution

Velocity. The x, y axes are located at an arbitrary fixed point. Sincethe motion relative to plane A is to be determined, the translatingframe of reference is attached to it, Fig. 12–44a. Applying therelative-velocity equation in scalar form since the velocity vectors ofboth planes are parallel at the instant shown, we have

Ans.

The vector addition is shown in Fig. 12–44b.

Acceleration. Plane B has both tangential and normal componentsof acceleration, since it is flying along a curved path. From Eq. 12–20,the magnitude of the normal component is

Applying the relative-acceleration equation, we have

Thus,

From Fig. 12–44c, the magnitude and direction of are therefore

Ans.

Notice that the solution to this problem is possible using atranslating frame of reference, since the pilot in plane A is“translating.” Observation of plane A with respect to the pilot of planeB, however, must be obtained using a rotating set of axes attached toplane B. (This assumes, of course, that the pilot of B is fixed in therotating frame, so he does not turn his eyes to follow the motion ofA.) The analysis for this case is given in Example 16.21.

aB>A = 912 km>h2 u = tan-1 150900

= 9.46°cu

aB>AaB>A = 5900i - 150j6 km>h2

900i - 100j = 50j + aB>AaB = aA + aB>A

1aB2n =vB

2

r=1600 km>h22

400 km= 900 km>h2

vB>A = -100 km>h = 100 km>hp600 = 700 + vB>AvB = vA + vB>A1+q2

y¿x¿,

rB = 400 km.

E X A M P L E 12.26

400 km

600 km/h

100 km/h2

y

x

y'

4 km

A Bx'

700 km/h

50 km/h2

(a)

150 km/h2

900 km/h2

(c)

θaB/A

Fig. 12–44

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Fig. 12–45B

Fig. 12–45A


At the instant shown in Fig. 12–45 cars A and B are traveling withspeeds of 18 m/s and 12 m/s, respectively. Also at this instant, A hasa decrease in speed of and B has an increase in speed of Determine the velocity and acceleration of B with respect to A.

Solution

Velocity. The fixed x, y axes are established at a point on the groundand the translating axes are attached to car A, Fig. 12–45a.Why?The relative velocity is determined from What arethe two unknowns? Using a Cartesian vector analysis, we have

Thus,

Ans.

Noting that has and components, Fig. 12–45b, its direction is

Ans.

Acceleration. Car B has both tangential and normal components ofacceleration. Why? The magnitude of the normal component is

Applying the equation for relative acceleration yields

Here has and components. Thus, from Fig. 12–45c,

Ans.

Ans.

It is possible to obtain the relative acceleration of using thismethod? Refer to the comment made at the end of Example 12.26.

aA>Bf = 62.7°d

tan f =1aB>A2y1aB>A2x =

4.7322.440

aB>A = 412.44022 + 14.73222 = 5.32 m>s2

-j- iaB>AaB>A = 5-2.440i - 4.732j6 m>s2

1-1.440i - 3j2 = 12 cos 60°i + 2 sin 60°j2 + aB>AaB = aA + aB>A

1aB2n =vB

2

r=112 m>s22

100 m= 1.440 m>s2

u = 21.7°a

tan u =1vB>A2y1vB>A2x =

3.5889

+j+ ivB>AvB>A = 21922 + 13.58822 = 9.69 m>s

vB>A = 59i + 3.588j6 m>s-12j = 1-18 cos 60°i - 18 sin 60°j2 + vB>AvB = vA + vB>A

vB = vA + vB>A.y¿x¿,

3 m>s2.2 m>s2,

E X A M P L E 12.27

y'

x'

y

x

= 100 mρ

12 m /s

3 m /s2

B

A

18 m /s

2 m /s2

60°

60°

(a)

(b)

9 m/s

3.588 m/s

θ

vB/A

(c)

4.732 m/s2

2.440 m/s2

φ

aB/A

Fig. 12–45

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*12-176. Determine the displacement of the log if thetruck at C pulls the cable 4 ft to the right.

12-177. The crate is being lifted up the inclined planeusing the motor M and the rope and pulley arrangementshown. Determine the speed at which the cable must betaken up by the motor in order to move the crate up theplane with a constant speed of 4 ft/s.

CB

Prob. 12–176

B

AC

M

Prob. 12–177


12-174. Determine the constant speed at which thecable at A must be drawn in by the motor in order tohoist the load at B 15 ft in 5 s.

12-175. Determine the time needed for the load at B toattain a speed of 8 m/s, starting from rest, if the cable isdrawn into the motor with an acceleration of 0.2 m>s2.

A

vA

vBB

A

vA

vBB

Prob. 12–174

Prob. 12–175

P R O B L E M S

*12-172. If the end of the cable at A is pulled down witha speed of 2 m/s,determine the speed at which block B rises.

12-173. If the end of the cable at A is pulled down with aspeed of 2 m/s, determine the speed at which block B rises.

A

B

C

2 m/s

Prob. 12–172

CD

2 m/sA

B

Prob. 12–173

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PROBLEMS • 89

C

D

B 4 ft/s

A

Prob. 12–179

B

A

CP

4 ft/s

Prob. 12–180

B

A

C

Prob. 12–178

12-178. Determine the displacement of the block at Bif A is pulled down 4 ft.

12-179. The cable at B is pulled downwards at 4 ft/s,and is slowing at Determine the velocity andacceleration of block A at this instant.

2 ft>s2.

12-181. If block A is moving downward with a speed of4 ft/s while C is moving up at 2 ft/s, determine the speedof block B.

12-182. If block A is moving downward at 6 ft/s whileblock C is moving down at 18 ft/s, determine the relativevelocity of block B with respect to C.

BC

AProbs. 12–181/182

12-183. If the end of the cable at A is pulled down with aspeed of 2 m/s, determine the speed at which block E rises.

*12-184. If block A of the pulley system is movingdownward with a speed of 4 ft/s while block C is movingup at 2 ft/s, determine the speed of block B.

B

C

D

E

2 m/s A

Prob. 12–183

A

B

C

Prob. 12–184

*12-180. The pulley arrangement shown is designed forhoisting materials. If BC remains fixed while the plunger Pis pushed downward with a speed of 4 ft/s, determine thespeed of the load at A.

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12-186. The cylinder C is being lifted using the cable andpulley system shown. If point A on the cable is beingdrawn toward the drum with a speed of 2 m/s, determinethe speed of the cylinder.

12-187. The motion of the collar at A is controlled by amotor at B such that when the collar is at it ismoving upwards at 2 ft/s and slowing down at Determine the velocity and acceleration of the cable asit is drawn into the motor B at this instant.

1 ft>s2.sA = 3 ft

*12-188. The roller at A is moving upward with avelocity of and has an acceleration of

when Determine the velocity andacceleration of block B at this instant.

sA = 4 ft.aA = 4 ft>s2vA = 3 ft>s

12-189. The crate C is being lifted by moving the rollerat A downward with a constant speed of along the guide. Determine the velocity and accelerationof the crate at the instant When the roller is atB, the crate rests on the ground. Neglect the size of thepulley in the calculation. Hint: Relate the coordinates and using the problem geometry, then take the firstand second time derivatives.

xA

xC

s = 1 m.

vA = 2 m>s

C

s

A

vA

Prob. 12–186

B

4 ft

A

sA

Prob. 12–187

B

A

3 ft sA = 4 ft

Prob. 12–188

s

C

A

4 m

4 m

B

xA

xC

Prob. 12–189


12-185. The crane is used to hoist the load. If the motorsat A and B are drawing in the cable at a speed of 2 ft/sand 4 ft/s, respectively, determine the speed of the load.

A B 4 ft/s2 ft/s

Prob. 12–185

12-190. The girl at C stands near the edge of the pierand pulls in the rope horizontally at a constant speed of6 ft/s. Determine how fast the boat approaches the pierat the instant the rope length AB is 50 ft.

AC

xC

xB

B

6 ft/s

8 ft

Prob. 12–190

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PROBLEMS • 91

12-191. The man pulls the boy up to the tree limb Cby walking backward. If he starts from rest when and moves backward with a constant acceleration

determine the speed of the boy at the instantNeglect the size of the limb. When so that A and B are coincident, i.e., the rope is

16 m long.yB = 8 m,

xA = 0,yB = 4 m.0.2 m>s2,

aA =xA = 0

8 m

yB

B

A

xA

C

Prob. 12–191

*12-192. Collars A and B are connected to the cord thatpasses over the small pulley at C. When A is located atD, B is 24 ft to the left of D. If A moves at a constantspeed of 2 ft/s to the right, determine the speed of B whenA is 4 ft to the right of D.

B

2 ft/s

A

D

C

10 ft

Prob. 12–192

12-193. If block B is moving down with a velocity and has an acceleration determine the velocity andacceleration of block A in terms of the parameters shown.

aB,vB

hA

B

vB, aB

sA

Prob. 12–193

12-194. Vertical motion of the load is produced bymovement of the piston at A on the boom. Determinethe distance the piston or pulley at C must move to theleft in order to lift the load 2 ft. The cable is attached atB, passes over the pulley at C, then D, E, F, and againaround E, and is attached at G.

12-195. Sand falls from rest 0.5 m vertically onto a chute.If the sand then slides with a velocity of downthe chute, determine the relative velocity of the sand justfalling on the chute at A with respect to the sand slidingdown the chute. The chute is inclined at an angle of 40°with the horizontal.

vC = 2 m>s

A6 ft/s

G

C

E

F

B

D

Prob. 12–194

0.5 m

= 40°θ

vA

vC = 2 m/sAC

Prob. 12–195

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12-197. At the instant shown, cars A and B are travelingat speeds of 30 mi/h and 20 mi/h, respectively. If B isincreasing its speed by while A maintains aconstant speed, determine the velocity and accelerationof B with respect to A.

12-198. At the instant shown, cars A and B are travelingat speeds of 30 mi/h and 20 mi/h, respectively. If A isincreasing its speed at whereas the speed of Bis decreasing at determine the velocity andacceleration of B with respect to A.

800 mi>h2,400 mi>h2

1200 mi>h2,

A

B0.3 mi

vB = 20 mi/h

vA = 30 mi/h

30°

Probs. 12–197/198


12-199. Two boats leave the shore at the same time andtravel in the directions shown. If and

determine the speed of boat A with respect to boatB. How long after leaving the shore will the boats be 800 ftapart?

15 ft>s,vB =vA = 20 ft>s

*12-200. Two planes A and B are flying side by side at aconstant speed of 900 km/h. Maintaining this speed, planeA begins to travel along the spiral path where is in radians, whereas plane B continues to fly in astraight line. Determine the speed of plane A with respectto plane B when r = 750 km.

u

r = 11500u2 km,

vA

vB

A

O

B

30°

45°

Prob. 12–199

θθ

r

r = 1500

A

B

Prob. 12–200

12-201. At the instant shown, the bicyclist at A is travelingat 7 m/s around the curve on the race track while increasinghis speed at The bicyclist at B is traveling at 8.5m/s along the straight-a-way and increasing his speed at

Determine the relative velocity and relativeacceleration of A with respect to B at this instant.0.7 m>s2.

0.5 m>s2.

12-202. At the instant shown, cars A and B are travelingat speeds of 55 mi/h and 40 mi/h, respectively. If B isincreasing its speed by while A maintains aconstant speed, determine the velocity and accelerationof B with respect to A. Car B moves along a curve havinga radius of curvature of 0.5 mi.

1200 mi>h2,

50 m 50 m

40°

vB = 8.5 m/svA = 7 m/s

A

B

Prob. 12–201

vB = 40 mi/h

vA = 55 mi/h

B

A 30°

Prob. 12–202

*12-196. Two planes, A and B, are flying at the samealtitude. If their velocities are and

such that the angle between theirstraight-line courses is determine the velocity ofplane B with respect to plane A.

u = 75°,vB = 500 km>h vA = 600 km>h

θ

A

B

vA

vBProb. 12–196

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PROBLEMS • 93

*12-204. The two cyclists A and B travel at the sameconstant speed v. Determine the speed of A with respectto B if A travels along the circular track, while B travelsalong the diameter of the circle.

v

v

rφ θ

A

B

Prob. 12–204

12-203. Cars A and B are traveling around the circularrace track. At the instant shown, A has a speed of 90 ft/sand is increasing its speed at the rate of whereasB has a speed of 105 ft/s and is decreasing its speed at

Determine the relative velocity and relativeacceleration of car A with respect to car B at this instant.25 ft>s2.

15 ft>s2,

A

B

vA

vB

rB = 250 ft

rA = 300 ft60°

Prob. 12–203

12-205. A man can row a boat at 5 m/s in still water.He wishes to cross a 50-m-wide river to point B, 50 mdownstream. If the river flows with a velocity of 2 m/s,determine the speed of the boat and the time needed tomake the crossing.

50 m

B

A

50 m

2 m/s

θ

Prob. 12–205

12-206. A passenger in an automobile observes thatraindrops make an angle of 30° with the horizontal as theauto travels forward with a speed of 60 km/h. Computethe terminal (constant) velocity of the rain if it isassumed to fall vertically.

vr

vr

va = 60 km/h

Prob. 12–206

12-207. At a given instant the football player at A throwsa football C with a velocity of 20 m/s in the direction shown.Determine the constant speed at which the player at B mustrun so that he can catch the football at the same elevationat which it was thrown. Also calculate the relative velocityand relative acceleration of the football with respect to B atthe instant the catch is made. Player B is 15 m away fromA when A starts to throw the football.

15 m

AB

C20 m/s

60°

Prob. 12–207

*12-208. A man can swim at 4 ft/s in still water. Hewishes to cross the 40-ft-wide river to point B, 30 ftdownstream. If the river flows with a velocity of 2 ft/s,determine the speed of the man and the time needed tomake the crossing. Note: While in the water he must notdirect himself towards point B to reach this point. Why?

B

A

30 ft

vr = 2 ft/s40 ft

Prob. 12–208

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D E S I G N P R O J E C T

12–1D. DESIGN OF A MARBLE-SORTING DEVICE

Marbles roll off the production chute at 0.5 ft/s. Determine the range forthe angle for a selected position s for the placement of thehopper relative to the end of the chute. Submit a drawing of the devicethat shows the path the marbles take.

0 … u … 30°3 ft

2 ft

s

0.5 ft/sθ

Fig. 12–1D

CHAPTER REVIEW• Rectilinear Kinematics. Rectilinear kinematics refers to motion along a straight line. A position

coordinate s specifies the location of the particle on the line, and the displacement is the change inthis position.

The average velocity is a vector quantity, defined as the displacement divided by the time interval.

This is different than the average speed, which is a scalar and is the total distance traveled divided by thetime of travel.

The time,position, instantaneous velocity,and instantaneous acceleration are related by the differential equations

If the acceleration is known to be constant, then integration of these equations yields

• Graphical Solutions. If the motion is erratic, then it can be described by a graph. If one of these graphsis given, then the others can be established using the differential relations, or

For example, if the v–t graph is known, then values of the s–t graph are determined fromincrements under the v–t graph. Values of the a–t graph are determined from

of v–t graph.

• Curvilinear Motion, x, y, z. For this case, motion along the path is resolved into rectilinear motion alongthe x, y, z axes. The equation of the path is used to relate the motion along each axis.

a = dv>dt = slope¢s = 1 v dt = areaa ds = v dv.

a = dv>dt,v = ds>dt,

v2 = v02 + 2ac1s - s02s = s0 + v0t + 1

2 act2

v = v0 + act

v = ds>dt a = dv>dt a ds = v dv

1vsp2avg =sT

¢t

vavg =¢r¢t

¢s

HIBBMC12_0131416782_2_95_v2 8/19/03 11:37 AM Page 94

• Projectile Motion. Free flight motion of a projectile follows a parabolic path. It has a constant velocityin the horizontal direction and constant acceleration of or in the vertical direction.Any two of the three equations for constant acceleration apply in the vertical direction, and in thehorizontal direction only applies.

• Curvilinear Motion n, t. If normal and tangential axes are used for the analysis, then v is always in thepositive t direction. The acceleration has two components. The tangential component, accounts for thechange in the magnitude of the velocity; a slowing down is in the negative t direction, and a speeding upis in the positive t direction. The normal component accounts for the change in the direction of thevelocity. This component is always in the positive n direction.

• Curvilinear Motion, r, z. If the path of motion is expressed in polar coordinates, then the velocity andacceleration components can be written as

To apply these equations, it is necessary to determine at the instant considered. If the pathis given, then the chain rule of calculus must be used to obtain the time derivatives. Once the

data is substituted into the equations, then the algebraic sign of the results will indicate the direction ofthe components of or a along each axis.

• Absolute Dependent Motion of Two Particles. The dependent motion of blocks that are suspended frompulleys and cables can be related by the geometry of the system. This is done by first establishing positioncoordinates, measured from a fixed origin to each block so that they are directed along the line of motionof the blocks. Using geometry and/or trigonometry, the coordinates are then related to the cable lengthin order to formulate a position coordinate equation. The first time derivative of this equation gives arelationship between the velocities of the blocks, and a second time derivative gives the relationshipbetween their accelerations.

• Relative Motion Analysis Using Translating Axes. If two particles A and B undergo independentmotions, then these motions can be related to their relative motion. Using a translating set of axes attachedto one of the particles (A), the velocity and acceleration equations become

For planar motion, each of these equations produces two scalar equations, one in the x, and the other in they direction. For solution, the vectors can be expressed in Cartesian form or the x and y scalar components canbe written directly.

aB = aA + aB>AvB = vA + vB>A

v

r = f1u2 r,.r,

..r,

.u,

..u

vu = r.u au = r

..u + 2

.r

.u

vr = r#

ar = ..r - r

.u2

U,

an

at,

x = x0 + 1v02xt

32.2 ft>s2g = 9.81 m>s2

CHAPTER REVIEW • 95

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Hibbeler dynamics ch12

Mobile