Kinematics

Unit 2 Physics: Kinamatics: M J Rhoades      

Kinematics (from Greek κινεῖν, kinein, to move) is the branch of classical mechanics that describes the motion of bodies (objects) and systems (groups of objects) without consideration of the forces that cause the motion.

Kinematics is not to be confused with another branch of classical mechanics: analytical dynamics (the study of the relationship between the motion of objects and its causes), sometimes subdivided into kinetics (the study of the relation between external forces and motion) and statics (the study of the relations in a system at equilibrium).Kinematics also differs from dynamics as used in modern-day physics to describe time-evolution of a system.

The Kinematic Equations:The goal of this first unit of The Physics Classroom has been to investigate the variety of means by which the motion of objects can be described. The variety of representations that we have investigated includes verbal representations, pictorial representations, numerical representations, and graphical representations (position-time graphs and velocity-time graphs). In Lesson 6, we will investigate the use of equations to describe and represent the motion of objects. These equations are known as kinematic equations.

There are a variety of quantities associated with the motion of objects - displacement (and distance), velocity (and speed), acceleration, and time. Knowledge of each of these quantities provides descriptive information about an object's motion. For example, if a car is known to move with a constant velocity of 22.0 m/s, north for 12.0 seconds for a northward displacement of 264 meters, then the motion of the car is fully described. And if a second car is known to accelerate from a rest position with an eastward acceleration of 3.0 m/s2 for a time of 8.0 seconds, providing a final velocity of 24 m/s, East and an eastward displacement of 96 meters, then the motion of this car is fully described. These two statements provide a complete description of the motion of an object. However, such completeness is not always known. It is often the case that only a few parameters of an object's motion are known, while the rest are unknown. For example as you approach the stoplight, you might know that your car has a velocity of 22 m/s, east and is capable of a skidding acceleration of 8.0 m/s2, West. However you do not know the displacement that your car would experience if you were to slam on your brakes and skid to a stop; and you do not know the time required to skid to a stop. In such an instance as this, the unknown parameters can be determined using physics principles and mathematical equations (the kinematic equations).

The kinematic equations are a set of four equations that can be utilized to predict unknown information about an object's motion if other information is known. The equations can be utilized for any motion that can be described as being either a constant velocity motion (an acceleration of 0 m/s/s) or a constant acceleration motion. They can never be used over any time period

during which the acceleration is changing. Each of the kinematic equations includes four variables. If the values of three of the four variables are known, then the value of the fourth variable can be calculated. In this manner, the kinematic equations provide a useful means of predicting information about an object's motion if other information is known. For example, if the acceleration value and the initial and final velocity values of a skidding car are known, then the displacement of the car and the time can be predicted using the kinematic equations. Lesson 6 of this unit will focus upon the use of the kinematic equations to predict the numerical values of unknown quantities for an object's motion.

The four kinematic equations that describe an object's motion are: See the cheat sheet in doc-sharing for derivation.

There are a variety of symbols used in the above equations. Each symbol has its own specific meaning. The symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol stands for the acceleration of the object. And the symbol v stands for the velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value.

Each of these four equations appropriately describes the mathematical relationship between the parameters of an object's motion. As such, they can be used to predict unknown information about an object's motion if other information is known. In the next part of Lesson 6 we will investigate the process of doing this.

The term kinematics is less common today than in the past, but still has a role in physics. The term kinematics also finds use in biomechanics and animal locomotion. Further, mathematicians that include time as a parameter in geometry have developed the subject of kinematic geometry.

The simplest application of kinematics is for particle motion, translational or rotational. The next level of complexity comes from the introduction of rigid bodies, which are collections of particles having time invariant distances between themselves. Rigid bodies might undergo translation and rotation or a combination of both. A more complicated case is the kinematics of a system of rigid bodies, which may be linked together by mechanical joints. Kinematics can be used to find the possible range of

motion for a given mechanism, or, working in reverse, can be used to design a mechanism that has a desired range of motion. The movement of a crane and the oscillations of a piston in an engine are both simple kinematic systems. The crane is a type of open kinematic chain, while the piston is part of a closed four-bar linkage.

Linear motion:

Linear or translational kinematics is the description of the motion in space of a point along a line, also known as a trajectory or path. This path can be either straight (rectilinear) or curved (curvilinear).

Particle kinematics

Particle kinematics is the study of the kinematics of a single particle. The results obtained in particle kinematics are used to study the kinematics of collection of particles, dynamics and in many other branches of mechanics. Position and reference frames

The position of a point in space is the most fundamental idea in particle kinematics. To specify the position of a point, one must specify three things: the reference point (often called the origin), distance from the reference point and the direction in space of the straight line from the reference point to the particle. Exclusion of any of these three parameters renders the description of position incomplete. Consider for example a tower 50 m south from your home. The reference point is home, the distance 50 m and the direction south. If one only says that the tower is 50 m south, the natural question that arises is "from where?" If one says that the tower is southward from your home, the question that arises is "how far?" If one says the tower is 50 m from your home, the question that arises is "in which direction?" Hence, all these three parameters are crucial to defining uniquely the position of a point in space.

Position is usually described by mathematical quantities that have all these three attributes: the most common are vectors and complex numbers. Usually, only vectors are used. For measurement of distances and directions, usually three dimensional coordinate systems are used with the origin coinciding with the reference point. A three-dimensional coordinates system (whose origin coincides with the reference point) with some provision for time measurement is called a reference frame or frame of reference or simply frame. All observations in physics are incomplete without the reference frame being specified.

Position vector

The position vector of a particle is a vector drawn from the origin of the reference frame to the particle. It expresses both the distance of the point from the origin and its sense from the origin. In three dimensions, the position of point A can be expressed as

where xA, yA, and zA are the Cartesian coordinates of the point. The magnitude of the position vector |r| gives the distance between the point A and the origin.

The direction cosines of the position vector provide a quantitative measure of direction. It is important to note that the position vector of a particle isn't unique. The position vector of a given particle is different relative to different frames of reference.

Rest and motion

Once the concept of position is firmly established, the ideas of rest and motion naturally follow. If the position vector of the particle (relative to a given reference frame) changes with time, then the particle is said to be in motion with respect to the chosen reference frame. However, if the position vector of the particle (relative to a given reference frame) remains the same with time, then the particle is said to be at rest with respect to the chosen frame. Note that rest and motion are relative to the reference frame chosen. It is quite possible that a particle at rest relative to a particular reference frame is in motion relative to the other. Hence, rest and motion aren't absolute terms; rather they are dependent on reference frame. For example, a passenger in a moving car may be at rest with respect to the car, but in motion with respect to the road.

Path

A particle's path is the locus between its beginning and end points which is reference-frame dependent. The path of a particle may be rectilinear (straight line) in one frame, and curved in another.

Displacement

Displacement is a vector describing the difference in position between two points, i.e. it is the change in position the particle undergoes during the time interval. If point A has position rA = (xA,yA,zA) and point B has position rB = (xB,yB,zB), the displacement rAB of B from A is given by

Geometrically, displacement is the shortest distance between the points A and B. Displacement, distinct from position vector, is independent of the reference frame. This can be understood as follows: the positions of points are frame dependent, however, the shortest distance between any pair of points is invariant on translation from one frame to another (barring relativistic cases).

The distance traveled is always greater than or equal to the displacement.

Distance

Distance is a scalar quantity, describing the length of the path between two points along which a particle has travelled.

When considering the motion of a particle over time, distance is the length of the particle's path and may be different from displacement, which is the change from its initial position to its final position. For example, a race car traversing a 10 km closed loop from start to finish travels a distance of 10 km; its displacement, however, is zero because it arrives back at its initial position.

Velocity and speed

Average velocity is defined as

Where Δr is the change in position and Δt is the interval of time over which the position changes. The direction of v is same as the direction of the change in position Δr as Δt>0.

Velocity is the measure of the rate of change in position with respect to time, that is, how the distance of a point changes with each instant of time. Velocity also is a vector. Instantaneous velocity (the velocity at an instant of time) can be defined as the limiting value of average velocity as the time interval Δt becomes smaller and smaller. Both Δr and Δt approach zero but the ratio v approaches a non-zero limit v. That is,

Where dr is an infinitesimally small displacement and dt is an infinitesimally small length of time. As per its definition in the derivative form, velocity can be said to be the time rate of change of position. Further, as dr is tangential to the actual path, so is the velocity.

As a position vector itself is frame dependent, velocity is also dependent on the reference frame.

The speed of an object is the magnitude |v| of its velocity. It is a scalar quantity:

The distance traveled by a particle over time is a non-decreasing quantity. Hence, ds/dt is non-negative, which implies that speed is also non-negative.

Acceleration

Average acceleration (acceleration over a length of time) is defined as:

where Δv is the change in velocity and Δt is the interval of time over which velocity changes.

Acceleration is the vector quantity describing the rate of change with time of velocity. Instantaneous acceleration (the acceleration at an instant of time) is defined as the limiting value of average acceleration as Δt becomes smaller and smaller. Under such a limit, a → a.

where dv is an infinitesimally small change in velocity and dt is an infinitesimally small length of time.

Types of motion based on velocity and acceleration

If the acceleration of a particle is zero, then the velocity of the particle is constant over time and the motion is said to be uniform. Otherwise, the motion is non-uniform.

If the acceleration is non-zero but constant, the motion is said to be motion with constant acceleration. On the other hand, if the acceleration is variable, the motion is called motion with variable acceleration. In motion with variable acceleration, the rate of change of acceleration is called the jerk.

Kinematics of constant acceleration

Many physical situations can be modeled as constant-acceleration processes, such as projectile motion.

Integrating acceleration a with respect to time t gives the change in velocity.

A relationship without explicit time dependence may also be derived for one-dimensional motion. Noting that at = v − v0,

where · denotes the dot product. Dividing the t on both sides and carrying out the dot-products:

In the case of straight-line motion, (r - r0) is parallel to a. Then

This relation is useful when time is not known explicitly.

Relative velocity

To describe the motion of object A with respect to object B, when we know how each is moving with respect to a reference object O, we can use vector algebra. Choose an origin

for reference, and let the positions of objects A, B, and O be denoted by rA, rB, and rO. Then the position of A relative to the reference object O is

Consequently, the position of A relative to B is

The above relative equation states that the motion of A relative to B is equal to the motion of A relative to O minus the motion of B relative to O. It may be easier to visualize this result if the terms are re-arranged:

or, in words, the motion of A relative to the reference is that of B plus the relative motion of A with respect to B. These relations between displacements become relations between velocities by simple time-differentiation, and a second differentiation makes them apply to accelerations.

For example, let Ann move with velocity   relative to the reference (we drop

the O subscript for convenience) and let Bob move with velocity , each velocity given with respect to the ground (point O). To find how fast Ann is

moving relative to Bob (we call this velocity ), the equation above gives:

To find   we simply rearrange this equation to obtain:

At velocities comparable to the speed of light, these equations are not valid. They are replaced by equations derived from Einstein's theory of special relativity.

Kinematics is the study of how things move. Here, we are interested in the motion of normal objects in our world. A normal object is visible, has edges, and has a location that can be expressed with (x, y, z) coordinates. We will not be discussing the motion of atomic particles or black holes or light.

We will create a vocabulary and a group of mathematical methods that will describe this ordinary motion. Understand that we will be developing a language for describing motion only. We won't be concerned with what is causing or changing the motion, or more

correctly, the momentums of the objects. In other words, we are not concerned with the action of forces within this topic.

Rotational motion

Circular motion

The angular velocity vector Ω points up for counterclockwise rotation and down for clockwise rotation, as specified by the right-hand rule. Angular position θ (t) changes with time at a rate ω (t) = dθ/dt.

Rotational or angular kinematics is the description of the rotation of an object. The description of rotation requires some method for describing orientation, for example, In what follows; attention is restricted to simple rotation about an axis of fixed orientation. The z-axis has been chosen for convenience.

Description of rotation then involves these three quantities:

Angular position: The oriented distance from a selected origin on the rotational axis to a point of an object is a vector r (t) locating the point. The vector r (t) has

some projection (or, equivalently, some component) r⊥ (t) on a plane perpendicular to the axis of rotation. Then the angular position of that point is the

angle θ from a reference axis (typically the positive x-axis) to the vector r⊥ (t) in a known rotation sense (typically given by the right).

Angular velocity: The angular velocity ω is the rate at which the angular position θ changes with respect to time t:

The angular velocity is represented by a vector Ω pointing along the axis of rotation with magnitude ω and sense determined by the direction of rotation as given by the right-hand rule.

Angular acceleration: The magnitude of the angular acceleration α is the rate at which the angular velocity ω changes with respect to time t:

The equations of translational kinematics can easily be extended to planar rotational kinematics with simple variable exchanges

See the cheat sheet in doc-sharing for the derivation of the kinematic formulas

Figure 2: Velocity and acceleration for nonuniform circular motion: the velocity vector is tangential to the orbit, but the acceleration vector is not radially inward because of its tangential component aθ that increases the rate of rotation: dω/dt = |aθ|/R.

This example deals with a "point" object, by which is meant that complications due to rotation of the body itself about its own center of mass are ignored.

Displacement. An object in circular motion is located at a position r (t) given by:

where uR is a unit vector pointing outward from the axis of rotation toward the periphery of the circle of motion, located at a radius R from the axis.

Linear velocity. The velocity of the object is then

The magnitude of the unit vector uR (by definition) is fixed, so its time dependence is entirely due to its rotation with the radius to the object, that is,

where uθ is a unit vector perpendicular to uR pointing in the direction of rotation, ω (t) is the (possibly time varying) angular rate of rotation, and the symbol × denotes the vector cross product. The velocity is then:

The velocity therefore is tangential to the circular orbit of the object, pointing in the direction of rotation, and increasing in time if ω increases in time.

Linear acceleration. In the same manner, the acceleration of the object is defined as:

which shows a leading term aθ in the acceleration tangential to the orbit related to the angular acceleration of the object (supposing ω to vary in time) and a second term aR directed inward from the object toward the center of rotation, called the centripetal acceleration.

[edit]Coordinate systems

See also: Generalized coordinates, Curvilinear coordinates, Orthogonal coordinates, and Frenet-Serret formulas

In any given situation, the most useful coordinates may be determined by constraints on the motion, or by the geometrical nature of the force causing or affecting the motion. Thus, to describe the motion of a bead constrained to move along a circular hoop, the most useful coordinate may be its angle on the hoop. Similarly, to describe the motion of a particle acted upon by a central force, the most useful coordinates may be polar coordinates. Polar coordinates are extended into three dimensions with either the spherical polar or cylindrical polar coordinate systems. These are most useful in systems exhibiting spherical or cylindrical symmetry respectively.

[edit]Fixed rectangular coordinates

In this coordinate system, vectors are expressed as an addition of vectors in the x, y, and z direction from a non-rotating origin. Usually i, j, k are unit vectors in thex-, y-, and z-directions.

The position vector, r, the velocity vector, v, and the acceleration vector,  are expressed using rectangular coordinates in the following way:

Note:   , 

[edit]Two-dimensional rotating reference frame

See also: Centripetal force

This coordinate system expresses only planar motion. It is based on three orthogonal unit vectors: the vector i, and the vector j which form a basis for the plane in which the objects we are considering reside, and k about which rotation occurs. Unlike rectangular coordinates, which are measured relative to an origin that is fixed and non-rotating, the origin of these coordinates can rotate and translate - often following a particle on a body that is being studied.

[edit]Derivatives of unit vectors

The position, velocity, and acceleration vectors of a given point can be expressed using these coordinate systems, but we have to be a bit more careful than we do with fixed frames of reference. Since the frame of reference is rotating, the unit vectors also rotate, and this rotation must be taken into account when taking the derivative of any of these vectors. If the coordinate frame is rotating at angular rate ω in the counterclockwise direction (that is, Ω = ω k using the right hand rule) then the derivatives of the unit vectors are as follows:

[edit]Position, velocity, and acceleration

Given these identities, we can now figure out how to represent the position, velocity, and acceleration vectors of a particle using this reference frame.

[edit]Position

Position is straightforward:

It is just the distance from the origin in the direction of each of the unit vectors.

[edit]Velocity

Velocity is the time derivative of position:

By the product rule, this is:

Which from the identities above we know to be:

or equivalently

where vrel is the velocity of the particle relative to the rotating coordinate system.

[edit]Acceleration

Acceleration is the time derivative of velocity.

We know that:

Consider the   part.   has two parts we want to find the derivative of: the relative change in velocity ( ), and the change in the coordinate frame

( ).

Next, consider  . Using the chain rule:

 from above:

So all together:

And collecting terms:[10]

[edit]Kinematic constraints

A kinematic constraint is any condition relating properties of a dynamic system that must hold true at all times. Below are some common examples:

[edit]Rolling without slipping

An object that rolls against a surface without slipping obeys the condition that the velocity of its center of mass is equal to the cross product of its angular velocity with a vector from the point of contact to the center of mass,

.

For the case of an object that does not tip or turn, this reduces to v = R ω.

[edit]Inextensible cord

This is the case where bodies are connected by an idealized cord that remains in tension and cannot change length. The constraint is that the sum of lengths of all segments of the cord is the total length, and accordingly the time derivative of this sum is zero. A dynamic problem of this type is the pendulum. Another example is a drum turned by the pull of gravity upon a falling weight attached to the rim by the inextensible cord.

In mathematics, a line refers to a straight trajectory, and a curve to a trajectory which may have curvature. In mechanics and kinematics, "line' and "curve" both refer to any trajectory, in particular a line may be a complex curve in space. Any position along a specified trajectory can be described by a single coordinate, the distance traversed along the path, or arc length. The motion of a particle along a trajectory can be described by specifying the time dependence of its position, for example by specification of the arc length locating the particle at each time t. The following words refer to curves and lines:

"linear" (= along a straight or curved line;

"rectilinear" (= along a straight line, from Latin rectus = straight, and linere = spread),

"curvilinear" (=along a curved line, from Latin curvus = curved, and linere = spread).

^ Because magnitude of dr is necessarily the distance between two infinitesimally spaced points along the trajectory of the point, it is the same as an increment in arc length along the path of the point, customarily denoted ds.

Interaction

In physics, acceleration is the rate of change of velocity with time. In one dimension,

acceleration is the rate at which something speeds up or slows down. However, since velocity is

a vector, acceleration describes the rate of change of both the magnitude and the direction of

velocity.[2][3] Acceleration has the dimensions L T −2. In SI units, acceleration is measured

in meters per second squared (m/s2). (Negative acceleration i.e. retardation, also has the same

dimensions/units.)

Proper acceleration, the acceleration of a body relative to a free-fall condition, is measured by an

instrument called an accelerometer.

In common speech, the term acceleration is used for an increase in speed (the magnitude of

velocity); a decrease in speed is called deceleration. In physics, a change in the direction of

velocity also is an acceleration: for rotary motion, the change in direction of velocity results

in centripetal (toward the center) acceleration; whereas the rate of change of speed is a tangential

acceleration.

In classical mechanics, for a body with constant mass, the acceleration of the body is

proportional to the net forceacting on it (Newton's second law):

where F is the resultant force acting on the body, m is the mass of the body, and a is its

acceleration.

Acceleration is the rate of change of velocity. At any point on a trajectory, the magnitude

of the acceleration is given by the rate of change of velocity in both magnitude and

direction at that point. The true acceleration at time t is found in the limit as time

interval Δt → 0.

Components of acceleration for a planar curved motion. The tangential component at is

due to the change in speed of traversal, and points along the curve in the direction of the

velocity vector. The centripetal component ac is due to the change in direction of the

velocity vector and is normal to the trajectory, pointing toward the center of curvature of

the path.

Average acceleration is the change in velocity (Δv) divided by the change in time (Δt).

Instantaneous acceleration is the acceleration at a specific point in time which is for a very

short interval of time as Δt approaches zero.

velocity: is the measurement of the rate and direction of change in the position of an

object. It is a vector physical quantity; both magnitude and direction are required to

define it. The scalar absolute value (magnitude) of velocity is speed, a quantity that is

measured in meters per second (m/s or ms−1) when using the SI (metric) system.

For example, "5 meters per second" is a scalar and not a vector, whereas "5 meters per

second east" is a vector. The average velocity v of an object moving through

a displacement   during a time interval (Δt) is described by the formula:

The rate of change of velocity is acceleration – how an object's speed or direction of travel

changes over time, and how it is changing at a particular point in time.

a. Example problems An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.

See solution below.

b. A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.

c. Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.6 seconds, what will be his final velocity and how far will he fall?

See solution below.

d. A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled.

See solution below.

e. A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s2. Determine the time for the feather to fall to the surface of the moon.

See solution below.

f. Rocket-powered sleds are used to test the human response to acceleration. If a rocket-powered sled is accelerated to a speed of 444 m/s in 1.8 seconds, then what is the acceleration and what is the distance that the sled travels?

See solution below.

g. A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4 m. Determine the acceleration of the bike.

See solution below.

h. An engineer is designing the runway for an airport. Of the planes that will use the airport, the lowest acceleration rate is likely to be 3 m/s2. The takeoff speed for this plane will be 65 m/s. Assuming this minimum acceleration, what is the minimum allowed length for the runway?

See solution below.

i. A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the skidding distance of the car (assume uniform acceleration).

See solution below.

j. A kangaroo is capable of jumping to a height of 2.62 m. Determine the takeoff speed of the kangaroo.

See solution below.

k. If Michael Jordan has a vertical leap of 1.29 m, then what is his takeoff speed and his hang time (total time to move upwards to the peak and then return to the ground)?

See solution below.

l. A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 0.840 m. Determine the acceleration of the bullet (assume a uniform acceleration).

See solution below.

m. A baseball is popped straight up into the air and has a hang-time of 6.25 s. Determine the height to which the ball rises before it reaches its peak. (Hint: the time to rise to the peak is one-half the total hang-time.)

See solution below.

n. The observation deck of tall skyscraper 370 m above the street. Determine the time required for a penny to free fall from the deck to the street below.

See solution below.

o. A bullet is moving at a speed of 367 m/s when it embeds into a lump of moist clay. The bullet penetrates for a distance of 0.0621 m. Determine the acceleration of the bullet while moving into the clay. (Assume a uniform acceleration.)

p. A stone is dropped into a deep well and is heard to hit the water 3.41 s after being dropped. Determine the depth of the well.

q. It was once recorded that a Jaguar left skid marks that were 290 m in length. Assuming that the Jaguar skidded to a stop with a constant acceleration of -3.90 m/s2, determine the speed of the Jaguar before it began to skid.

r. A plane has a takeoff speed of 88.3 m/s and requires 1365 m to reach that speed. Determine the acceleration of the plane and the time required to reach this speed.

s. A dragster accelerates to a speed of 112 m/s over a distance of 398 m. Determine the acceleration (assume uniform) of the dragster.

t. With what speed in miles/hr (1 m/s = 2.23 mi/hr) must an object be thrown to reach a height of 91.5 m (equivalent to one football field)? Assume negligible air resistance.

Solutions to Above

a. : Given:

a = +3.2 m/s2 t = 32.8 s vi = 0 m/s

Find:

d = ??

b. d = vi*t + 0.5*a*t2

c. d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s2)*(32.8 s)2

d. d = 1720 m

e.

Given:

d = 110 m t = 5.21 s vi = 0 m/s

Find:

a = ??

f. d = vi*t + 0.5*a*t2

g. 110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s)2

h. 110 m = (13.57 s2)*a

i. a = (110 m)/(13.57 s2)

j. a = 8.10 m/ s2

Given:

a = -9.8 m t = 2.6 s vi = 0 m/s

Find:

d = ??

vf = ??

k. d = vi*t + 0.5*a*t2

l. d = (0 m/s)*(2.6 s)+ 0.5*(-9.8 m/s2)*(2.6 s)2

m. d = -33 m (- indicates direction)

n. vf = vi + a*t

o. vf = 0 + (-9.8 m/s2)*(2.6 s)

p. vf = -25.5 m/s (- indicates direction)

Given: Find:

vi = 18.5 m/s vf = 46.1 m/s t = 2.47 s d = ??

a = ??

q. a = (Delta v)/t

r. a = (46.1 m/s - 18.5 m/s)/(2.47 s)

s. a = 11.2 m/s2

t. d = vi*t + 0.5*a*t2

u. d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s2)*(2.47 s)2

v. d = 45.7 m + 34.1 m

w. d = 79.8 m

x. (Note: the d can also be calculated using the equation vf2 = vi

2 + 2*a*d)

Given:

vi = 0 m/s d = -1.40 m a = -1.67 m/s2

Find:

t = ??

y. d = vi*t + 0.5*a*t2

z. -1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s2)*(t)2

aa. -1.40 m = 0+ (-0.835 m/s2)*(t)2

bb. (-1.40 m)/(-0.835 m/s2) = t2

cc. 1.68 s2 = t2

dd. t = 1.29 s

ee.

Given:

vi = 0 m/s vf = 44 m/s t = 1.80 s

Find:

a = ??

d = ??

ff. a = (Delta v)/t

gg. a = (444 m/s - 0 m/s)/(1.80 s)

hh. a = 247 m/s2

ii. d = vi*t + 0.5*a*t2

jj. d = (0 m/s)*(1.80 s)+ 0.5*(247 m/s2)*(1.80 s)2

kk. d = 0 m + 400 m

ll. d = 400 m

mm. (Note: the d can also be calculated using the equation vf2 = vi

2 + 2*a*d)

nn.

Given:

vi = 0 m/s vf = 7.10 m/s d = 35.4 m

Find:

a = ??

oo. vf2 = vi

2 + 2*a*d

pp. (7.10 m/s)2 = (0 m/s)2 + 2*(a)*(35.4 m)

qq. 50.4 m2/s2 = (0 m/s)2 + (70.8 m)*a

rr. (50.4 m2/s2)/(70.8 m) = a

ss. a = 0.712 m/s2

tt.

Given:

vi = 0 m/s vf = 65 m/s a = 3 m/s2

Find:

d = ??

uu. vf2 = vi

2 + 2*a*d

vv. (65 m/s)2 = (0 m/s)2 + 2*(3 m/s2)*d

ww. 4225 m2/s2 = (0 m/s)2 + (6 m/s2)*d

xx. (4225 m2/s2)/(6 m/s2) = d

yy. d = 704 m

zz.

Given:

vi = 22.4 m/s vf = 0 m/s t = 2.55 s

Find:

d = ??

aaa. d = (vi + vf)/2 *t

bbb. d = (22.4 m/s + 0 m/s)/2 *2.55 s

ccc. d = (11.2 m/s)*2.55 s

ddd. d = 28.6 m

eee.

Given:

a = -9.8 m/s2 vf = 0 m/s d = 2.62 m

Find:

vi = ??

fff. vf2 = vi

2 + 2*a*d

ggg. (0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(2.62 m)

hhh. 0 m2/s2 = vi2 - 51.35 m2/s2

iii. 51.35 m2/s2 = vi2

jjj. vi = 7.17 m/s

Given:

a = -9.8 m/s2 vf = 0 m/s d = 1.29 m

Find:

vi = ??

t = ??

kkk. vf2 = vi

2 + 2*a*d

lll. (0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(1.29 m)

mmm. 0 m2/s2 = vi2 - 25.28 m2/s2

nnn. 25.28 m2/s2 = vi2

ooo. vi = 5.03 m/s

ppp. To find hang time, find the time to the peak and then double it.

qqq. vf = vi + a*t

rrr. 0 m/s = 5.03 m/s + (-9.8 m/s2)*tup

sss. -5.03 m/s = (-9.8 m/s2)*tup

ttt. (-5.03 m/s)/(-9.8 m/s2) = tup

uuu. tup = 0.513 s

vvv. hang time = 1.03 s

www.

Given:

vi = 0 m/s vf = 521 m/s d = 0.840 m

Find:

a = ??

xxx. vf2 = vi

2 + 2*a*d

yyy. (521 m/s)2 = (0 m/s)2 + 2*(a)*(0.840 m)

zzz. 271441 m2/s2 = (0 m/s)2 + (1.68 m)*a

aaaa. (271441 m2/s2)/(1.68 m) = a

bbbb. a = 1.62*105 m /s2

Given:

a = -9.8 m/s2 vf = 0 m/s t = 3.13 s

Find:

d = ??

a. (NOTE: the time required to move to the peak of the trajectory is one-half the total hang time.)

First use: vf = vi + a*t

0 m/s = vi + (-9.8 m/s2)*(3.13 s)

0 m/s = vi - 30.6 m/s

vi = 30.6 m/s

Now use: vf2 = vi

2 + 2*a*d

(0 m/s)2 = (30.6 m/s)2 + 2*(-9.8 m/s2)*(d)

0 m2/s2 = (938 m/s) + (-19.6 m/s2)*d

-938 m/s = (-19.6 m/s2)*d

(-938 m/s)/(-19.6 m/s2) = d

d = 47.9 m

cccc.

Given:

vi = 0 m/s d = -370 m a = -9.8 m/s2

Find:

t = ??

dddd. d = vi*t + 0.5*a*t2

eeee. -370 m = (0 m/s)*(t)+ 0.5*(-9.8 m/s2)*(t)2

ffff. -370 m = 0+ (-4.9 m/s2)*(t)2

gggg. (-370 m)/(-4.9 m/s2) = t2

hhhh. 75.5 s2 = t2

iiii. t = 8.69 s

Given:

vi = 367 m/s vf = 0 m/s d = 0.0621 m

Find:

a = ??

jjjj. vf2 = vi

2 + 2*a*d

kkkk. (0 m/s)2 = (367 m/s)2 + 2*(a)*(0.0621 m)

llll. 0 m2/s2 = (134689 m2/s2) + (0.1242 m)*a

mmmm. -134689 m2/s2 = (0.1242 m)*a

nnnn. (-134689 m2/s2)/(0.1242 m) = a

oooo. a = -1.08*106 m /s2

pppp. (The - sign indicates that the bullet slowed down.)

qqqq.

Given:

a = -9.8 m/s2 t = 3.41 s vi = 0 m/s

Find:

d = ??

rrrr. d = vi*t + 0.5*a*t2

ssss. d = (0 m/s)*(3.41 s)+ 0.5*(-9.8 m/s2)*(3.41 s)2

tttt. d = 0 m+ 0.5*(-9.8 m/s2)*(11.63 s2)

uuuu. d = -57.0 m

vvvv. (NOTE: the - sign indicates direction)

wwww.  

xxxx.

Given:

a = -3.90 m/s2 vf = 0 m/s d = 290 m

Find:

vi = ??

yyyy. vf2 = vi

2 + 2*a*d

zzzz. (0 m/s)2 = vi2 + 2*(-3.90 m/s2)*(290 m)

aaaaa. 0 m2/s2 = vi2 - 2262 m2/s2

bbbbb. 2262 m2/s2 = vi2

ccccc.vi = 47.6 m /s

ddddd.  

eeeee.

Given:

vi = 0 m/s vf = 88.3 m/s d = 1365 m

Find:

a = ??

t = ??

fffff. vf2 = vi

2 + 2*a*d

ggggg. (88.3 m/s)2 = (0 m/s)2 + 2*(a)*(1365 m)

hhhhh. 7797 m2/s2 = (0 m2/s2) + (2730 m)*a

iiiii. 7797 m2/s2 = (2730 m)*a

jjjjj. (7797 m2/s2)/(2730 m) = a

kkkkk. a = 2.86 m/s2

lllll. vf = vi + a*t

mmmmm. 88.3 m/s = 0 m/s + (2.86 m/s2)*t

nnnnn. (88.3 m/s)/(2.86 m/s2) = t

ooooo. t = 30. 8 s

ppppp.  

qqqqq.

Given:

vi = 0 m/s vf = 112 m/s d = 398 m

Find:

a = ??

rrrrr. vf2 = vi

2 + 2*a*d

sssss. (112 m/s)2 = (0 m/s)2 + 2*(a)*(398 m)

ttttt. 12544 m2/s2 = 0 m2/s2 + (796 m)*a

uuuuu. 12544 m2/s2 = (796 m)*a

vvvvv. (12544 m2/s2)/(796 m) = a

wwwww. a = 15.8 m/s2

xxxxx.  

yyyyy.

Given:

a = -9.8 m/s2 vf = 0 m/s d = 91.5 m

Find:

vi = ??

t = ??

zzzzz.First, find speed in units of m/s:

aaaaaa. vf2 = vi

2 + 2*a*d

bbbbbb. (0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(91.5 m)

cccccc. 0 m2/s2 = vi2 - 1793 m2/s2

dddddd. 1793 m2/s2 = vi2

eeeeee. vi = 42.3 m/s

ffffff. Now convert from m/s to mi/hr:

gggggg. vi = 42.3 m/s * (2.23 mi/hr)/(1 m/s)

hhhhhh. vi = 94.4 mi/hr