Chp11-p36-52

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36

PROBLEM 11.31

The acceleration due to gravity of a particle falling toward the earth is 2 2/ ,a gR r= − where r is the distance from the center of the earth to the

particle, R is the radius of the earth, and g is the acceleration due to

gravity at the surface of the earth. If 6370R = km, calculate the escapevelocity, that is, the minimum velocity with which a particle must be

projected upward from the surface of the earth if it is not to return to

earth. (Hint: 0v = for .r = ∞ )

SOLUTION

The acceleration is given by2

2

dv gRv a

dr r= = −

Then,

2

2

gR drv dv

r= −

Integrating, using the conditions esc0 at , and v r v v= = ∞ = at r R=

esc

0 2

2v R

drv dv gR

r

∞= −∫ ∫

esc

02 21 1

2 v R

v gRr

∞ =

2 2esc

1 10 0

2v gR

R

− = −

esc 2v gR=

3 2Now, 6370 km 6370 10 m and 9.81 m/s .R g= = × =

Then, ( )( )( )esc 2v = 2esc 111.8 10 m/s v = × �9.81 36370 10×




37

PROBLEM 11.32

The acceleration due to gravity at an altitude y above the surface of the

earth can be expressed as

( ) 26

32.2

1 / 20.9 10

a

y

−= + ×

where a and y are expressed in 2m/s and feet, respectively. Using this

expression, compute the height reached by a projectile fired vertically

upward from the surface of the earth if its initial velocity is

(a) 720 1200 12,000v = m/s, (b) v = m/s, (c) v = m/s.

SOLUTION

The acceleration is given by

6

2

20.9 10

32.2

1y

a

×

−= + 6

2

20.9 10

32.2

1y

dyvdv ady

×

−= = +

20 maxIntegrate, using the conditions at 0 and 0 at . Also, use 9.81 m/s andv v y v y y g= = = = =

36370 10 m.R = ×

( ) ( )0

0 2

2 20 0

1v

y

R

dy dyv dv g gR

R y

∞ ∞= − = −++

∫ ∫ ∫max

0

02 2

0

1 1

2

y

v

v gRR y

= +

( )2 2 2max0 0 max max

max max

1 1 10 2

2

gRyv gR v R y gRy

R y R R y

− = − = − + = + +

maxSolving for ,y20

max 202

Rvy

gR v=

−

Using the given numerical data,( )( )( )

2 20 0

max 22002

v vy

vv= =

−−

0( ) 720 m/s,a v =( )( )

( ) ( )

2

max 2720

y =−

max

y = �

0( ) 1200 m/s,b v =( )( )

( ) ( )

2

max 2

1200y =

−max 74250 m y = �

0( ) 12,000 m/s,c v =( )( )

( ) ( )

2

max 2negativey = =

−

Negative value indicates that 0v is greater than the escape velocity.

maxy = ∞ �

4637 10

×36370 10×36370 10× 4 12498 10×9.81

412498 10×

4637 10× 72026532 m

4637 10×

1200412498 10×

4637 10×412498 10×

12,000

12,000

where a and y are expressed in m/s2 and m, respectively. Using this

,




38

PROBLEM 11.33

The velocity of a slider is defined by the relation 'sin( ).nv v tω ϕ= +Denoting the velocity and the position of the slider at 0t = by 0v and

0,x respectively, and knowing that the maximum displacement of the

slider is 02 ,x show that (a) ( )2 2 20 0 0' / 2n nv v x xω ω= + , (b) the maximum

value of the velocity occurs when 20 0 03 ( / ) / 2.nx x v x w = −

SOLUTION

( )( ) Given: sin na v v tω ϕ′= +

At 0,t = 00 sin or sin

vv v v

vϕ ϕ′= = =

′ (1)

Let x be maximum at 1t t= when 0.v =

Then, ( ) ( )1 1sin 0 and cos 1n nt tω ϕ ω ϕ+ = + = ± (2)

Using or dx

v dx v dtdt

= =

Integrating, ( )cos nn

vx C tω ϕ

ω′

= − +

At 0,t = 0 0cos or cosn n

v vx x C C xϕ ϕ

ω ω′ ′

= = − = +

Then, ( )0 cos cos nn n

v vx x tϕ ω ϕ

ω ω′ ′

= + − + (3)

max 0 1cos using cos 1nn

v vx x tϕ ω ϕ

ω ω′ ′

= + + + = −

Solving for cos ,ϕ ( )max 0cos 1

nx x

v

ωϕ

−= −

′

max 0With 2 ,x x= 0cos 1nx

v

ωϕ = −′ (4)

Using

2 22 2 0 0sin cos 1, or 1 1nv x

v v

ωϕ ϕ + = + − = ′ ′

Solving for givesv′( )2 2 2

0 0

0

(5) 2

n

n

v xv

x

ω

ω

+′ = t




39

PROBLEM 11.33 CONTINUED

( ) Acceleration:b ( )cosn n

dva v t

dtω ω ϕ′= = +

2Let be maximum at when 0.v t t a= =

Then, ( )2cos 0ntω ϕ+ =

From equation (3), the corresponding value of x is

( )

00 0 0

2 2 2 20 0 0

0 0 20 0

cos 1 2

3 12

2 2 2

n

n n n

n

n n n

v v x vx x x x

v

v x vx x

x x

ωϕω ω ω

ωω ω ω

′ ′ ′ = + = + − = − ′

+= − = −

( )0

0

2

0

3

2

n

v

x

xω

− t

From Equation (3), the corresponding value of x is




40

PROBLEM 11.34

The velocity of a particle is ( )0 1 sin / .v v t Tπ = − Knowing that the

particle starts from the origin with an initial velocity 0v , determine (a) its

position and its acceleration at 3 ,t T= (b) its average velocity during the

interval 0t = to .t T=

SOLUTION

0( ) 1 sindx t

a v vdt T

π = = −

0Integrating, using 0 when 0,x x t= = =

00 0 01 sin

x t t tdx v dt v dt

T

π = = − ∫ ∫ ∫

000

0

cos

tx v T t

x v tT

ππ

= +

0 00 cos

v T t v Tx v t

T

ππ π

= + − (1)

When 3 ,t T= ( )0 00 0

23 cos 3 3

v T v Tx v T v T

Tπ

π π = + − = −

02.36 x v T= t

0 cosdv v t

adt T T

π π= = −

When 3 ,t T= 0 cos3v

aT

π π= − 0va

T

π= t

( ) Using equation (1) with ,b t T=

0 01 0 0

2cos 1

v T v Tx v T v Tπ

π π π = + − = −

Average velocity is

1 0ave 0

21

x x xv v

t T π∆ − = = = − ∆ ave 00.363 v v= t

Using Equation (1) with t = T,




41

PROBLEM 11.35

A minivan is tested for acceleration and braking. In the street-start

acceleration test, elapsed time is 8.2 s for a velocity increase from

10 km/h to 100 km/h. In the braking test, the distance traveled is 44 m

during braking to a stop from 100 km/h. Assuming constant values of

acceleration and deceleration, determine (a) the acceleration during the

street-start test, (b) the deceleration during the braking test.

SOLUTION

10 km/h 2.7778 m/s= 100 km/h 27.7778 m/s=

(a) Acceleration during start test.

dva

dt= 8.2 27.7778

0 2.7778a dt v dt=∫ ∫

8.2 27.7778 2.7778a = − 23.05 m/sa = t

(b) Deceleration during braking.

dva v

dx= =

44 0

0 27.7778a dx v dv= =∫ ∫

( ) ( )0

44 2

027.7778

1

2a x v=

( )2144 27.7778

2a = −

28.77 m/sa = − deceleration 28.77 m/sa= − = t

m/s,

,

braking:




42

PROBLEM 11.36

In Prob. 11.35, determine (a) the distance traveled during the street-start

acceleration test, (b) the elapsed time for the braking test.

SOLUTION

10 km/h 2.7778 m/s= 100 km/h 27.7778 m/s=

(a) Distance traveled during start test.

dva

dt=

00

t v

va dt dv=∫ ∫

0at v v= − 0v va

t

−=

227.7778 2.77783.04878 m/s

8.2a

−= =

0 2.7778 3.04878v v at t= + = +

( )8.2

0 02.7778 3.04878

tx v dv t dt= = +∫ ∫

( )( ) ( )( )22.7778 8.2 1.52439 8.2= + 125.3 mx = �

(b) Elapsed time for braking test.

dva v

dx=

00

x v

va dx v dv=∫ ∫

2 20

2 2

v vax = −

( ) ( )( ) ( )2 2 20

1 10 27.7778

2 2 44a v v

x= − = −

28.7682 m/s= −

dva

dt=

00

t v

va dt dv=∫ ∫

0at v v= −

0 0 27.7778

8.7682

v vt

a

− −= =−

3.17 st = �

m/s,

test:

,

,

test:

,

,




43

PROBLEM 11.37

An airplane begins its take-off run at A with zero velocity and a constant

acceleration a. Knowing that it becomes airborne 30 s later at B and that

the distance AB is 823 m, determine (a) the acceleration a, (b) the take-

off velocity .Bv

SOLUTION

Constant acceleration. 0 00, 0A Av v x x= = = =

0v v at at= + = (1)

2 20 0

1 1

2 2x x v t at at= + + = (2)

At point ,B 823 m and 30 sBx x t= = =

(a) Solving (2) for a,( )( )

( )2 2

2 8232

30

xa

t= = 2m/sa = t

(b) Then, ( )( )1.8 30Bv at= = 54 m/s Bv = t

1.8

Constant acceleration:




44

PROBLEM 11.38

Steep safety ramps are built beside mountain highways to enable vehicles

with defective brakes to stop safely. A truck enters a 229 m ramp at a

high speed 0v and travels 165 m in 6 s at constant deceleration before its

speed is reduced to 0 / 2.v Assuming the same constant deceleration,

determine (a) the additional time required for the truck to stop, (b) the

additional distance traveled by the truck.

SOLUTION

Constant acceleration. 0 0x =

0v v at= + (1)

20 0

1

2x x v t at= + + (2)

Solving (1) for a,0v v

at

−= (3)

Then, ( ) ( )200 0 0 0 0

1 1 1

2 2 2

v vx x v t t x v v t v v t

t

−= + + = + + = +

At 6 s,t = 0 6

1 and 165 m

2v v x= =

( )0 0 0 0

0

1 1 165165 6 4.5 or 36.7 m/s

2 2 4.5

118.3 m/s

2

v v v v

v v

= + = = =

= =

Then, from (3), 2 218.3 36.7m/s 3.05 m/s

6 6a

−= = − = −

Substituting into (1) and (2), 36.7 3.05v t= −

( ) 210

2x t t= + −

At stopping, 0 or 0 12 ss sv t t= − = =

( )( ) ( )( )210 12 12 220.8 m

2x = + − =

( ) Additional time for stopping 12 s 6 sa = − 6 s t∆ = t

( ) Additional distance for stopping 220.8 m 165 mb = − 55.8 md∆ = t

18.3

36.7 3.05

36.7 3.05

36.7 3.05

229-m


0,




45

PROBLEM 11.39

A sprinter in a 400-m race accelerates uniformly for the first 130 m and

then runs with constant velocity. If the sprinter’s time for the first 130 m

is 25 s, determine (a) his acceleration, (b) his final velocity, (c) his time

for the race.

SOLUTION

20 0

1( ) During the acceleration phase

2a x x v t at= + +

0 0Using 0, and 0, and solving for givesx v a= =

2

2xa

t=

Noting that 130 m when 25 s,x t= =

( )( )( )2

2 130

25a = 0.416 m/s a = t

(b) Final velocity is reached at 25 s.t =

( )( )0 0 0.416 25fv v at= + = + 10.40 m/s fv = t

(c) The remaining distance for the constant speed phase is

400 130 270 mx∆ = − =

For constant velocity,270

25.96 s10.40

xt

v

∆∆ = = =

Total time for run: 25 25.96t = + 51.0 s t = t




46

PROBLEM 11.40

A group of students launches a model rocket in the vertical direction.

Based on tracking data, they determine that the altitude of the rocket was

27.5 m at the end of the powered portion of the flight and that the rocket

landed 16 s later. Knowing that the descent parachute failed to deploy so

that the rocket fell freely to the ground after reaching its maximum

altitude and assuming that 29.81 m/s ,g = determine (a) the speed 1v of

the rocket at the end of powered flight, (b) the maximum altitude reached

by the rocket.

SOLUTION

Constant acceleration. Choose 0t = at end of powered flight.

Then, 21 27.5 m 9.81 m/sy a g= = − = −

(a) When y reaches the ground, 0 and 16 s.fy t= =

2 21 1 1 1

1 1

2 2fy y v t at y v t gt= + + = + −

( )( )221 11 2 2

1

0 27.5 9.81 1676.76 m/s

16

fy y gtv

t

− + − += = =

1 76.8 m/s v = t

(b) When the rocket reaches its maximum altitude max,y

0v =

( ) ( )2 2 21 1 1 12 2v v a y y v g y y= + − = − −

2 21

12

v vy y

g

−= −

( )( )( )

2

max

0 76.7627.5

2 9.81y

−= − max 328 m y = t





47

PROBLEM 11.41

Automobile A starts from O and accelerates at the constant rate of

0.75 m/s2. A short time later it is passed by bus B which is traveling in the

opposite direction at a constant speed of 6 m/s. Knowing that bus Bpasses point O 20 s after automobile A started from there, determine

when and where the vehicles passed each other.

SOLUTION

Place origin at 0.

Motion of auto. ( ) ( ) 2

0 00, 0, 0.75 m/sA A Ax v a= = =

( ) ( ) ( )2 2

0 0

1 10 0 0.75

2 2A A A Ax x v t a t t

= + + = + +

20.375 mAx t=

Motion of bus. ( ) ( )0 0

?, 6 m/s, 0B B Bx v a= = − =

( ) ( ) ( )0 0 0

6 mB B B Bx x v t x t= − = −

At 20 , 0.Bt s x= =

( ) ( )( )0

0 6 20Bx= − ( )0

120 mBx =

Hence, 120 6Bx t= −

When the vehicles pass each other, .B Ax x=

2120 6 0.375t t− =

20.375 6 120 0t t+ − =

( )( )( )( )( )

26 (6) 4 0.375 120

2 0.375t

− ± − −=

6 14.69711.596 s and 27.6 s

0.75t

− ±= = −

Reject the negative root. 11.60 st = t

Corresponding values of xA and xB.

( )( )20.375 11.596 50.4 mAx = =

( )( )120 6 11.596 50.4 mBx = − = 50.4 mx = t

at 0:

,

0.75 m/s2. A short time later, it is passed by bus B which is traveling in the




48

PROBLEM 11.42

Automobiles A and B are traveling in adjacent highway lanes and at

0t = have the positions and speeds shown. Knowing that automobile Ahas a constant acceleration of 0.6 m/s

2and that B has a constant

deceleration of 0.4 m/s2, determine (a) when and where A will overtake

B, (b) the speed of each automobile at that time.

SOLUTION

Place the origin at A when t = 0.

Motion of A: ( ) ( ) 2

0 00, 15 km/h = 4.1667 m/s, 0.6 m/sA A Ax v a= = =

( )0

4.1667 0.6A A Av v a t t= + = +

( ) ( ) 2 2

0 0

14.1667 0.3

2A A A Ax x v t a t t t= + + = +

Motion of B: ( ) ( ) 2

0 025 m, 23 km/h = 6.3889 m/s, 0.4 m/sB B Bx v a= = = −

( )0

6.3889 0.4B B Bv v a t t= + = −

( ) ( ) 2 2

0 0

125 6.3889 0.2

2B B B Bx x v t a t t t= + + = + −

(a) When and where A overtakes B. A Bx x=

2 24.1667 0.3 25 6.3889 0.2t t t t+ = + −

20.5 2.2222 25 0t t− − =

( )( )( )( )( )

22.2222 2.2222 4 0.5 25

2 0.5t

± − −=

2.2222 7.4120 9.6343 s and 5.19 st = ± = −

Reject the negative root. . 9.63 st = t

( )( ) ( )( )24.1667 9.6343 0.3 9.6343 68.0 mAx = + =

( )( ) ( )( )225 6.3889 9.6343 0.2 9.6343 68.0 mBx = + − =

moves 68.0 mA t

moves 43.0 mB t

(b) Corresponding speeds.

( )( )4.1667 0.6 9.6343 9.947 m/sAv = + = 35.8 km/hAv = t

( )( )6.3889 0.4 9.6343 2.535 m/sBv = − = 9.13 km/hBv = t

B:

speeds:




49

PROBLEM 11.43

In a close harness race, horse 2 passes horse 1 at point A, where the two

velocities are 2v = 6.4 m/s and 1v = 6.2 m/s. Horse 1 later passes horse 2

at point B and goes on to win the race at point C, 366 m from A. The

elapsed times from A to C for horse 1 and horse 2 are 1t = 61.5 s and

2t = 62.0 s, respectively. Assuming uniform accelerations for both horses

between A and C, determine (a) the distance from A to B, (b) the position

of horse 1 relative to horse 2 when horse 1 reaches the finish line C.

SOLUTION

Constant acceleration ( )1 2 and a a for horses 1 and 2.

Let 0x = and 0t = when the horses are at point A.

Then, 20

1

2x v t at= +

Solving for , a( )0

2

2 x v ta

t

−=

Using 366 mx = and the initial velocities and elapsed times for each horse,

( )( )( )

–31 11 2 2

1

2 366 6.2 61.58 × 10 m/s

61.5

x v ta

t

−− = = = −

( )( )( )

22 22 2 2

2

2 6.4 62.01.6 10 m/s

62.0

x v ta

t

−− = = = −

1 2Calculating ,x x− ( ) ( ) 21 2 1 2 1 2

1

2x x v v t a a t− = − + −

( ) ( ) ( ) 21 2

2

1

2

.2 .004

x x t t

t t

− = − + − − −

= − +

At point B, 21 2 0 0B Bx x t t− = − + =

(a)0.2

50 s.004Bt = =

Calculating Bx using data for either horse,

Horse 1: ( )( ) ( )( )21

2Bx = + − 340 mBx = �

Horse 2: ( )( ) ( )( )21340 m

2Bx = + − =

When horse 1 crosses the finish line at 61.5 s,t =

(b) ( )( ) ( )( )2

1 2 61.5 61.5x x− = − + 2.8 mx∆ = �

− 2

366 − 2×

6.2 6.4 –38 × 10− 21.6 10 −×

.004

506.2 –38 × 10− 50

506.4 21.6 10 −× 50

0.2 .004

0.2

2:

366 m and the initial velocities and elapsed times for each horse,




50

PROBLEM 11.44

Two rockets are launched at a fireworks performance. Rocket A is

launched with an initial velocity 0v and rocket B is launched 4 s later

with the same initial velocity. The two rockets are timed to explode

simultaneously at a height of 73 m, as A is falling and B is rising.

Assuming a constant acceleration g = 9.81 m/s2 determine (a) the initial

velocity 0,v (b) the velocity of B relative to A at the time of the

explosion.

SOLUTION

Choose x positive upward. Constant acceleration a g= −

Rocket launch data: Rocket :A 00, , 0x v v t= = =

Rocket : B 00, , 4 sBx v v t t= = = =

Velocities: Rocket :A 0Av v gt= −

Rocket : B ( )0B Bv v g t t= − −

Positions: 20

1Rocket :

2AA x v t gt= −

( ) ( )2

0

1Rocket : ,

2B B B BB x v t t g t t t t= − − − ≥

For simultaneous explosions at 73 m when ,A B Ex x t t= = =

( ) ( )22 2 20 0 0 0

1 1 1 1

2 2 2 2E E E B E B E B E E B Bv t gt v t t g t t v t v t gt gt t gt− = − − − = − − + −

0Solving for , v 02

BE

gtv gt= − (1)

Then, when , Et t= 21,

2 2

BA E E E

gtx gt t gt

= − − or 2 2

0AE B E

xt t t

g− − =

Solving for , Et( )( )( ) ( ) ( )( )( )( )22 4 1 2 732

9.814 1 4 4

6.35 s2 2

AxB B g

E

t tt

± + ± += = =

Assuming a constant acceleration g = 9.81 m/s2, determine (a) the initial

t = tE,




51

PROBLEM 11.44 CONTINUED

(a) From equation (1), ( )( ) ( )( )0

9.81 49.81 6.35

2v = − 0 42.67 m/sv = �

At time ,Et 0A Ev v gt= − ( )0B E Bv v g t t= − −

(b) ( )( )9.81 4B A Bv v gt− = = / 39.24 m/sB Av = �

From Equation (1),

,




52

PROBLEM 11.45

In a boat race, boat A is leading boat B by 38 m and both boats are

traveling at a constant speed of 168 km/h. At 0t = , the boats accelerate

at constant rates. Knowing that when B passes A, 8t = s and

228Av = km/h, determine (a) the acceleration of A, (b) the acceleration

of B.

SOLUTION

(a) Acceleration of A.

( ) ( )0 0

, 168 km/h 46.67 m/sA A A Av v a t v= + = =

At 8 s,t = 228 km/h 63.33 m/sAv = =

( )0 63.33 46.67

8

A AA

v va

t

− −= = 22.08 m/s Aa = �

(b) ( ) ( ) 2

0 0

1

2A A A Ax x v t a t= + + ( ) ( ) 2

0 0

1

2B B B Bx x v t a t= + +

( ) ( ) ( ) ( ) ( ) 2

0 0 0 0

1

2A B A B A B A Bx x x x v v t a a t − = − + − + −

When 0,t = ( ) ( )0 0

38 mA Bx x− = and ( ) ( )0 0

0B Av v− =

When 8 s,t = 0A Bx x− =

Hence, ( )( )210 38 8 , or 1.1875

2A B A Ba a a a= + − − = −

1.1875 2.08 1.1875B Aa a= + = + 23.27 m/s Ba = �

of A:

at constant rates. Knowing that when B passes A, t = 8 s and

Chp11-p36-52

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