Chp11-p36-52

PROBLEM 11.31 The acceleration due to gravity of a particle falling toward the earth is a = − gR 2 / r 2 , 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 R = 6370 km, calculate the escape velocity, 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: v = 0 for r = ∞. )

SOLUTION dv gR 2 =a=− 2 dr r

The acceleration is given by

v

Then,

v dv = −

gR 2dr r2

Integrating, using the conditions v = 0 at r = ∞, and v = vesc at r = R dr

0 2 ∞ ∫ vesc v dv = − gR ∫ R r 2

1 2 v 2 0−

0 vesc

 1 = gR 2    r

∞

R

1 2 1  vesc = gR 2  0 −  2 R  vesc = 2 gR

Now, R = 6370 km = 6370 × 103 m and g = 9.81 m/s 2. Then,

vesc =

( 2 )(9.81) ( 6370 × 103)

vesc = 111.8 × 10 2 m/s �

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 36

PROBLEM 11.32 The acceleration due to gravity at an altitude y above the surface of the earth can be expressed as a=

− 32.2

(

)

1 + y / 20.9 × 106   

2

2 where a and yy are respectively. Using Using this this are expressed expressed inin m/s m/s2and andfeet, m, respectively. expression, compute the height reached by a projectile fired vertically upward from the surface of the earth if its initial velocity is (a) v = 720 m/s, (b) v = 1200 m/s, (c) v = 12,000 m/s.

SOLUTION The acceleration is given by a =

− 32.2   y  1 +  20.9 × 106  

2

− 32.2dy

vdv = ady =

,

  y  1 +  20.9 × 106  

2

Integrate, using the conditions v = v0 at y = 0 and v = 0 at y = ymax . Also, use g = 9.81 m/s 2 and R = 6370 × 103 m.

∫

0 v v0

0−

dv = − g ∫

∞ 0

(

dy

1+

y R

)

2

=

∞ − gR 2 0

∫

dy

( R + y )2

 1 2 1 1 gRymax − =− v0 = gR 2  + + ymax 2 R y R R max   ymax =

Using the given numerical data,

ymax =

6370 × 103 v02

( 2 )(9.81 ) (6370 × 103) − v02

( 637 × 10 )( 720) (12498 × 10 ) − ( 720 ) ( 637 × 10 )(1200 ) (12498 × 10 ) − (1200 ) ( 637 × 10 )(12,000 ) (12498 × 10 ) − (12,000)

(c) v0 = 12,000 m/s,

ymax =

ymax =

0

=

637 × 104 v02 2 12498 × 104 − v0

2

2

ymax = 26532 m �

4

2

ymax = 74250 m �

4

2

4

2

4

(b) v0 = 1200 m/s,

v0

ymax

v02 ( R + ymax ) = 2 gRymax

4

ymax =

1  = gR   R + y 

Rv02 2 gR − v02

Solving for ymax ,

(a) v0 = 720 m/s,

2

0

1 2 v 2

4

2

= negative

Negative value indicates that v0 is greater than the escape velocity. ymax = ∞ � PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 37

PROBLEM 11.33 The velocity of a slider is defined by the relation v = v 'sin(ω nt + ϕ ). Denoting the velocity and the position of the slider at t = 0 by v0 and x0 , respectively, and knowing that the maximum displacement of the

(

)

slider is 2 x0 , show that (a) v ' = v02 + x02ω n2 / 2 x0ω n , (b) the maximum value of the velocity occurs when x = x0 3 − (v0 / x0 wn ) 2  / 2.

SOLUTION (a) Given: v = v′ sin (ω nt + ϕ ) At t = 0,

v = v0 = v′ sin ϕ

sin ϕ =

or

v0 v′

(1)

Let x be maximum at t = t1 when v = 0. Then,

sin (ω nt1 + ϕ ) = 0

Using

dx =v dt

cos (ω nt1 + ϕ ) = ± 1

and

dx = v dt

or

v′ cos (ω nt + ϕ ) ωn

Integrating,

x=C−

At t = 0,

x = x0 = C −

Then,

x = x0 + xmax = x0 +

v′ cos ϕ ωn

C = x0 +

or

v′ cos ϕ ωn

v′ v′ cos ϕ − cos (ω nt + ϕ ) ωn ωn v′ v′ cos ϕ + ωn ω

Solving for cos ϕ ,

cos ϕ =

With xmax = 2 x0 ,

cos ϕ =

( xmax

(3)

using cos ω nt1 + ϕ = −1

− x0 )ω n −1 v′

x0ω n −1 v′

(4) 2

Using

(2)

sin 2 ϕ + cos 2ϕ = 1,

or

2

 v0   x0ω n   ′  +  ′ − 1 = 1 v   v  v′ =

Solving for v′ gives

(v

2 0

+ x02ω n2 2 x0ω n

)

(5) t


PROBLEM 11.33 CONTINUED a=

(b) Acceleration:

dv = v′ω n cos (ω nt + ϕ ) dt

Let v be maximum at t = t2 when a = 0. Then,

cos (ω nt2 + ϕ ) = 0

From equation Equation(3), (3),the thecorresponding correspondingvalue valueofofx is x is x = x0 +

v′ v′  x0ω n v′  cos ϕ = x0 +  ′ − 1 = 2 x0 − ωn ωn  v ωn 

= 2 x0 −

v02 + x02ω n2 3 1 v02 = x0 − 2 x0ω n2 (2 x0ω n )ω n 2  3 − x0 

( )  v0 x0ωn

2

2

t


PROBLEM 11.34 The velocity of a particle is v = v0 1 − sin (π t / T ) . Knowing that the particle starts from the origin with an initial velocity v0 , determine (a) its position and its acceleration at t = 3T , (b) its average velocity during the interval t = 0 to t = T .

SOLUTION (a )

dx πt   = v = v0 1 − sin  T  dt  Integrating, using x = x0 = 0 when t = 0,

πt

x t t   ∫ 0 dx = ∫ 0 v dt = ∫ 0 v0 1 − sin T  dt



x

x 0

vT πt   =  v0t + 0 cos  π T  

x = v0t + When t = 3T ,



t

0

v0T π t v0T cos − π π T

x = 3v0T +

(1)

v0T vT  2 cos (3π ) − 0 =  3 −  v0T T π π  x = 2.36v0T t

a= When t = 3T ,

dv πv πt = − 0 cos dt T T

a=−

π v0 cos 3π T

a=

π v0 t T

(b) Using equation , Equation(1) (1)with witht t==TT, x1 = v0T +

v0T vT 2  cos π − 0 = v0T 1 −  π π π 

Average velocity is vave =

∆x x1 − x0  2 = = 1 −  v0 ∆t T π 

vave = 0.363v0 t


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, m/s

100 km/h = 27.7778 m/s

(a) Acceleration during start test. a=

dv , dt

8.2 27.7778 ∫ 0 a dt = ∫ 2.7778 v dt

a = 3.05 m/s 2 t

8.2 a = 27.7778 − 2.7778 (b) Deceleration during braking. braking: a=v

dv = dx

44 0 ∫ 0 a dx = ∫ 27.7778 v dv =

a (x)

44 0

=

( )

1 2 v 2

44 a = −

0 27.7778

1 ( 27.7778)2 2 deceleration = − a = 8.77 m/s 2 t

a = − 8.77 m/s 2


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, m/s

100 km/h = 27.7778 m/s

(a) Distance traveled during start test: test. dv , dt

a=

t

v

∫ 0 a dt = ∫ v0 dv

at = v − v0 ,

a=

v − v0 t

27.7778 − 2.7778 = 3.04878 m/s 2 8.2

a=

v = v0 + at = 2.7778 + 3.04878 t x=

∫ 0 v dv = ∫ 0 (2.7778 + 3.04878 t ) dt t

8.2

= ( 2.7778)(8.2 ) + (1.52439 )(8.2 )

2

x = 125.3 m �

(b) Elapsed time for braking test: test. a=v

ax = a=

dv , dx

x

v

∫ 0 a dx = ∫ v0 v dv

v 2 v0 2 − 2 2

(

)

(

1 2 1 v − v02 = 0 − 27.77782 2x ( 2 )( 44 )

)

= − 8.7682 m/s 2 a=

dv , dt

t

v

∫ 0 a dt = ∫ v0 dv

at = v − v0 t=

v − v0 0 − 27.7778 = a − 8.7682

t = 3.17 s �


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 takeoff velocity vB .

SOLUTION Constant acceleration. Constant acceleration:

v0 = v A = 0,

x0 = x A = 0

v = v0 + at = at x = x0 + v0t +

1 2 1 2 at = at 2 2

At point B,

x = xB = 823 m

(a) Solving (2) for a,

a=

(b) Then,

(1)

2 x ( 2 )( 823 ) = t2 (30 )2

vB = at = (1.8)(30 )

and

(2) t = 30 s a = 1.8 m/s2 t vB = 54 m/s t


PROBLEM 11.38 Steep safety ramps are built beside mountain highways to enable vehicles 229-m with defective brakes to stop safely. A truck enters a 229 m ramp at a high speed v0 and travels 165 m in 6 s at constant deceleration before its speed is reduced to v0 / 2. 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 x0 = 0

Constant acceleration. Constant acceleration:

v = v0 + at

(1)

x = x0 + v0t + a=

Solving (1) for a, Then,

x = x0 + v0t +

At t = 6 s,

v= 165 = v=

1 v0 2

1 2 at 2

(2)

v − v0 t

(3)

1 v − v0 2 1 1 t = x0 + (v0 + v ) t = ( v0 + v ) t 2 t 2 2

and

x6 = 165 m

1 1   v0 + v0  (6 ) = 4.5v0 2 2 

or

v0 =

165 = 36.7 m/s 4.5

1 v0 = 18.3 m/s 2

Then, from (3),

18.3 − 36.7 18.3 2 =− a= m/s = − 3.05 m/s2 6 6

Substituting into (1) and (2),

v = 36.7 − 3.05t x = 0 + 36.7t −

At stopping, v = 0

or

0 36.7 − 3.05 ts = 0,

1 (3.05) t 2 2

ts = 12 s

x = 0 + ( 36.7 )(12 ) −

1 ( 3.05 )(12 )2 = 220.8 m 2

(a) Additional time for stopping = 12 s − 6 s (b) Additional distance for stopping = 220.8 m − 165 m

∆t = 6 s t

∆d = 55.8 m t


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 (a) During the acceleration phase x = x0 + v0t +

1 2 at 2

Using x0 = 0, and v0 = 0, and solving for a gives a=

2x t2

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

( 2 )(130 ) ( 25)2

a = 0.416 m/s t

(b) Final velocity is reached at t = 25 s. v f = v0 + at = 0 + ( 0.416 )( 25 )

v f = 10.40 m/s t

(c) The remaining distance for the constant speed phase is ∆x = 400 − 130 = 270 m For constant velocity, Total time for run:

∆t =

270 ∆x = = 25.96 s 10.40 v

t = 25 + 25.96

t = 51.0 s t


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 g = 9.81 m/s 2 , determine (a) the speed v1 of the rocket at the end of powered flight, (b) the maximum altitude reached by the rocket.

SOLUTION Constant acceleration. acceleration: Then,

Choose t = 0 at end of powered flight. y1 = 27.5 m

(a) When y reaches the ground, y f = 0

a = − g = − 9.81 m/s 2

y f = y1 + v1t +

v1 =

t = 16 s.

and

y f − y1 +

1 2 1 at = y1 + v1t − gt 2 2 2 1 2

t

gt 2

=

0 − 27.5 +

1 2

(9.81)(16 )2

16

= 76.76 m/s v1 = 76.8 m/s t

(b) When the rocket reaches its maximum altitude ymax , v=0 v 2 = v12 + 2a ( y − y1 ) = v12 − 2 g ( y − y1 ) y = y1 −

v 2 − v12 2g

0 − (76.76 ) = 27.5 − ( 2 )(9.81)

2

ymax

ymax = 328 m t


PROBLEM 11.41 Automobile A starts from O and accelerates at the constant rate of 0.75 m/s22.. A short time later, later ititis is passed passed by by bus bus B which is traveling in the opposite direction at a constant speed of 6 m/s. Knowing that bus B passes point O 20 s after automobile A started from there, determine when and where the vehicles passed each other.

SOLUTION Place origin at 0. 0:

( x A )0 = 0, (v A )0 = 0,

Motion of auto.

a A = 0.75 m/s 2

x A = ( x A )0 + ( v A )0 t +

1 1 a At 2 = 0 + 0 +   ( 0.75 ) t 2 2 2

x A = 0.375t 2 m Motion of bus.

( xB )0 = ?, (vB )0 = − 6 m/s,

aB = 0

xB = ( xB )0 − (vB )0 t = ( xB )0 − 6t m At t = 20 s, xB = 0. 0 = ( xB )0 − ( 6 )( 20 ) Hence,

( xB )0 = 120 m

xB = 120 − 6 t

When the vehicles pass each other, xB = x A. 120 − 6t = 0.375 t 2 0.375 t 2 + 6 t − 120 = 0 t= t=

− 6 ± (6) 2 − ( 4 )(0.375)( −120 )

( 2 )(0.375 )

− 6 ± 14.697 = 11.596 s 0.75

and − 27.6 s

Reject the negative root. Corresponding values of xA and xB,. x A = (0.375 )(11.596 ) = 50.4 m

t = 11.60 s t

2

xB = 120 − (6 )(11.596 ) = 50.4 m

x = 50.4 m t


PROBLEM 11.42 Automobiles A and B are traveling in adjacent highway lanes and at t = 0 have the positions and speeds shown. Knowing that automobile A has a constant acceleration of 0.6 m/s2 and 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:

( x A )0 = 0, (v A )0 = 15 km/h = 4.1667 m/s, v A = ( v A )0 + a At = 4.1667 + 0.6t

a A = 0.6 m/s 2

1 a At 2 = 4.1667 t + 0.3t 2 2 = 23 km/h = 6.3889 m/s, aB = − 0.4 m/s 2

x A = ( x A )0 + ( v A )0 t + Motion of B:

( xB )0 = 25 m, (vB )0 vB = ( vB )0 + aBt = 6.3889 − 0.4t xB = ( xB )0 + ( vB )0 t +

(a) When and where A overtakes B: B.

1 aBt 2 = 25 + 6.3889t − 0.2 t 2 2 x A = xB

4.1667 t + 0.3 t 2 = 25 + 6.3889 t − 0.2 t 2 0.5t 2 − 2.2222t − 25 = 0 t=

2.2222 ± 2.22222 − ( 4 )( 0.5 )( − 25 )

( 2 )(0.5)

t = 2.2222 ± 7.4120 = 9.6343 s and − 5.19 s

. t = 9.63 s t

Reject the negative root. x A = ( 4.1667 )(9.6343) + (0.3)(9.6343) = 68.0 m 2

xB = 25 + (6.3889 )(9.6343) − (0.2 )(9.6343) = 68.0 m 2

A moves 68.0 m t

B moves 43.0 m t

(b) Corresponding speeds: speeds.

v A = 35.8 km/h t

v A = 4.1667 + ( 0.6 )(9.6343) = 9.947 m/s

vB = 9.13 km/h t

vB = 6.3889 − (0.4 )(9.6343) = 2.535 m/s


PROBLEM 11.43 In a close harness race, horse 2 passes horse 1 at point A, where the two velocities are v2 = 6.4 m/s and v1 = 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 t1 = 61.5 s and t2 = 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 ( a1 and a2 ) for horses 1 and 2: 2. Let x = 0 and t = 0 when the horses are at point A. Then,

x = v0t +

Solving for a,

a=

1 2 at 2

2 ( x − v0t ) t2

Using x = 366 m and andthe theinitial initialvelocities velocitiesand andelapsed elapsedtimes timesfor foreach eachhorse, horse,

Calculating x1 − x2 ,

a1 =

x − v1t1 2  366 − (6.2 )( 61.5 ) = t12 (61.5)2

a2 =

x − v2t2 2 366 − ( 6.4 )( 62.0 ) = = − 1.6 × 10− 2 m/s 2 2 t22 62.0 ( ) x1 − x2 = (v1 − v2 ) t +

= − 8 ×10−–3 m/s 2

1 ( a1 − a2 ) t 2 2

x1 − x2 = ( 6.2 − 6.4) t +

1 ( − 8 ×10−–3 ) − ( − 1.6 × 10− 2 ) t 2 2

= − .2 t + .004 t 2 At point B, (a)

x1 − x2 = 0 tB =

− 0.2 t B + .004 tB2 = 0

0.2 = 50 s .004

Calculating xB using data for either horse, Horse 1: Horse 2:

xB = (6.2)(50) +

1 ( − 8 ×10−–3)( 50)2 2

xB = (6.4)(50) +

xB = 340 m�

1 ( − 1.6 × 10− 2 )(50)2 = 340 m 2

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

x1 − x2 = − ( 0.2)(61.5) + (.004)( 61.5)

2

∆x = 2.8 m �


PROBLEM 11.44 Two rockets are launched at a fireworks performance. Rocket A is launched with an initial velocity v0 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 determine (a) (a) the the initial initial Assuming aa constant constant acceleration acceleration g = 9.81 9.81 m/s m/s22, determine velocity v0 , (b) the velocity of B relative to A at the time of the explosion.

SOLUTION Choose x positive upward.

Constant acceleration a = − g Rocket A:

x = 0, v = v0 , t = 0

Rocket B:

x = 0, v = v0 , t = t B = 4 s

Rocket launch data:

Rocket A: v A = v0 − gt

Velocities:

Rocket B: vB = v0 − g (t − t B ) Rocket A: x A = v0t −

Positions:

1 2 gt 2

Rocket B: xB = v0 (t − t B ) −

1 2 g (t − t B ) , 2

t ≥ tB

For simultaneous explosions at x A = xB = 73 m when t = tEE,, v0t E −

1 2 1 1 1 2 gt E = v0 (t E − t B ) − g (t E − t B ) = v0t E − v0t B − gt E2 + gt E t B − gt B2 2 2 2 2 gt B 2

Solving for v0 ,

v0 = gt E −

Then, when t = t E ,

gt  1  x A =  gt E − B  t E − gt E2 , 2  2 

Solving for t E ,

tE =

(1)

t B ± t B2 + ( 4 )(1) 2

or

( ) = 4± 2 xA g

t E2 − t BtE −

2xA =0 g

2 )( 73) ( 4 )2 + (4)(1)(9.81

2

= 6.35 s


PROBLEM 11.44 CONTINUED (a) From From equation Equation(1), (1), At time t E , (b)

v0 = (9.81)( 6.35 ) −

(9.81 )(4 ) 2

v0 = 42.67 m/s �

vB = v0 − g (t E − t B )

v A = v0 − gt E ,

vB − v A = gt B = (9.81)(4 )

vB/ A = 39.24 m/s �


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 t = 0 , the boats accelerate at rates. Knowing Knowing that that when when BB passes at constant constant rates. passes A, A, t t == 88 ss and and v A = 228 km/h, determine (a) the acceleration of A, (b) the acceleration of B.

SOLUTION (a) Acceleration of A. A: v A = ( v A )0 + a At ,

(v A )0 = 168 km/h = 46.67 m/s v A = 228 km/h = 63.33 m/s

At t = 8 s, aA =

v A − ( v A )0 t

=

63.33 − 46.67 8

x A = ( x A )0 + ( v A )0 t +

(b)

1 a At 2 2

a A = 2.08 m/s 2 � xB = ( xB )0 + ( vB )0 t +

1 aB t 2 2

1 x A − xB = ( x A )0 − ( xB )0 + ( v A )0 − ( vB )0  t + ( a A − aB ) t 2 2 When t = 0,

( x A )0 − ( xB )0 = 38 m

When t = 8 s, Hence,

(vB )0 − (vA )0 = 0

and

x A − xB = 0 0 = 38 +

1 ( a A − aB )(8)2 , 2

or

aB = a A + 1.1875 = 2.08 + 1.1875

a A − aB = − 1.1875 aB = 3.27 m/s 2 �


Chp11-p36-52

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