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.
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 R=6370km, 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= for 0 r= ∞ ).
SOLUTION
The acceleration is given by
2 2 dv gR v a dr = = − r Then, 2 2 gR dr v dv r = −
Integrating, using the conditions v= 0 at r= ∞, and v=vesc at r =R
esc 0 2 2 v R dr v dv gR r ∞ = −
∫
∫
esc 0 2 2 1 1 2 v R v gR r ∞ = 2 2 esc 1 1 0 0 2v gR R − = − esc 2 v = gR 3 2 Now, R=6370 km =6370 10 m and × g =9.81 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.
37
PROBLEM 11.32
The acceleration due to gravity at an altitude y above the surface of the earth can be expressed as
(
)
2 6 32.2 1 / 20.9 10 a y − = + × wherea and y are expressed in m/s and feet, respectively. Using this 2 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
6 2 20.9 10 32.2 1 y a × − = + 6 2 20.9 10 32.2 1 y dy vdv ady × − = = + 2 0 max
Integrate, using the conditions v=v at y =0 and v=0 at y= y . Also, use g =9.81 m/s and
3 6370 10 m. R= ×
( )
(
)
0 0 2 2 2 0 0 1 v y R dy dy v dv g gR R y ∞ ∞ = − = − + +∫
∫
∫
max 0 0 2 2 0 1 1 2 y v v gR R y = + (
)
2 2 max 2 0 0 max max max max 1 1 1 0 2 2 gRy v gR v R y gRy R y R R y − = − = − + = + + max Solving for y , 2 0 max 2 0 2 Rv y gR v = − Using the given numerical data,( )(
)
(
)
2 2 0 0 max 2 2 0 0 2 v v y v v = = − − 0 ( ) a v =720 m/s,(
)
( )
(
)
(
)
2 max 2 720 y = − ymax = � 0 ( ) b v =1200 m/s,(
)
(
)
(
)
(
)
2 max 2 1200 y = − ymax =74250 m� 0 ( ) c v =12,000 m/s,(
)
(
)
(
)
(
)
2 max 2 negative y = = − Negative value indicates that v is greater than the escape velocity. 0max y = ∞ � 4 637 10× 3 6370 10× 3 6370 10× 12498 10× 4 9.81 4 12498 10× 4 637 10× 720 26532 m 4 637 10× 1200 4 12498 10× 4 637 10× 4 12498 10× 12,000 12,000
where a and y are expressed in m/s2 and m, respectively. Using this
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.
The velocity of a slider is defined by the relation v=v'sin(ωnt +ϕ). Denoting the velocity and the position of the slider at t = by 0 v and 0
0,
x respectively, and knowing that the maximum displacement of the slider is 2 ,x show that (a)0 v'=
(
v02 +x02ωn2)
/ 2x0ωn, (b) the maximum value of the velocity occurs when x=x03−(v0/x w0 n)2/ 2.SOLUTION
(
)
( ) Given: a v=v′sin ωnt +ϕ At t =0, v v0 vsin or sin v0 v ϕ ϕ ′ = = = ′ (1)Let x be maximum at t = when t1 v=0.
Then, sin
(
ωnt1+ϕ)
=0 and cos(
ωnt1 +ϕ)
= ± 1 (2)Using dx v or dx v dt dt = = Integrating, cos
(
n)
n v x C ω t ϕ ω ′ = − + At t =0, 0 cos or 0 cos n n v v x x C ϕ C x ϕ ω ω ′ ′ = = − = +Then, 0 cos cos
(
n)
n n v v x x ϕ ω t ϕ ω ω ′ ′ = + − + (3)
max 0 cos using cos n1 1
n v v x x ϕ ω t ϕ ω ω ′ ′ = + + + = −
Solving for cos ,ϕ cos
(
xmax x0)
n 1v ω ϕ = −′ − max 0 With x = 2 ,x cos x0 n 1 v ω ϕ = − ′ (4) Using 2 2 2 2 0 0 sin cos 1, or v x n 1 1 v v ω ϕ + ϕ = ′ + ′ − =
Solving for givesv′
(
)
2 2 2 0 0 0 (5) 2 n n v x v x ω ω + ′ = t
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.
39
PROBLEM 11.33 CONTINUED
( ) Acceleration:b a dv v ncos(
nt)
dt ′ω ω ϕ = = + 2Let be maximum at v t =t when a=0.
Then, cos
(
ωnt2 +ϕ)
=0From equation (3), the corresponding value of x is
(
)
0 0 0 0 2 2 2 2 0 0 0 0 0 2 0 0 cos 1 2 3 1 2 2 2 2 n n n n n n n n v v x v x x x x v v x v x x x x ω ϕ ω ω ω ω ω ω ω ′ ′ ′ = + = + ′ − = − + = − = −( )
0 0 2 0 3 2 n v x x ω − tPROPRIETARY 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.
The velocity of a particle is v=v01−sin
(
πt T/)
Knowing that the . particle starts from the origin with an initial velocity v , determine (a) its 0 position and its acceleration at t =3 ,T (b) its average velocity during the interval t = to 0 t =T. SOLUTION 0 ( ) a dx v v 1 sin t dt T π = = − 0Integrating, using x=x =0 when t =0,
0 0 0 0 1 sin x t t t dx v dt v dt T π = = −
∫
∫
∫
0 0 0 0 cos t x v T t x v t T π π = + 0 0 0 cos v T t v T x v t T π π π = + − (1) When t =3 ,T x 3v T0 v T0 cos 3( )
v T0 3 2 v T0 T π π π = + − = − 0 2.36 x= v T t 0cos dv v t a dt T T π π = = − When t =3 ,T a v0cos 3 T π π = − a v0 T π = t( ) Using equation (1) with b t =T,
0 0 1 0 0 2 cos 1 v T v T x v T π v T π π π = + − = − Average velocity is 1 0 ave 0 2 1 x x x v v t T π ∆ − = = = − ∆ vave =0.363 v0t
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.
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.
dv a dt = 8.2 27.7778 0 a dt = 2.7778 v dt
∫
∫
8.2a=27.7778−2.7778 a=3.05 m/s2t(b) Deceleration during braking. dv a v dx = = 44 0 0 a dx= 27.7778v dv=
∫
∫
( )
44( )
2 0 0 27.7778 1 2 a x = v(
)
2 1 44 27.7778 2 a= − 2 8.77 m/s a= − deceleration= − =a 8.77 m/s2t m/s, , braking: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.
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.
dv a dt = 0 0 t v v a dt = dv
∫
∫
0 at = −v v a v v0 t − = 2 27.7778 2.7778 3.04878 m/s 8.2 a= − = 0 2.7778 3.04878 v=v +at = + t(
)
8.2 0 0 2.7778 3.04878 t x =∫
v dv=∫
+ t dt(
)( ) (
)( )
2 2.7778 8.2 1.52439 8.2 = + x=125.3 m�(b) Elapsed time for braking test. dv a v dx = 0 0 x v v a dx= v dv
∫
∫
2 2 0 2 2 v v ax= −(
2 2)
( )( )
(
2)
0 1 1 0 27.7778 2 2 44 a v v x = − = − 2 8.7682 m/s = − dv a dt = 0 0 t v v a dt = dv∫
∫
0 at = −v v 0 0 27.7778 8.7682 v v t a − − = = − t =3.17 s� m/s, test: , , test: , ,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.
43
PROBLEM 11.37
An airplane begins its take-off run at A with zero velocity and a constant accelerationa. 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 vB. SOLUTION Constant acceleration. v0=vA =0, x0 =xA=0 0 v=v +at=at (1) 2 2 0 0 1 1 2 2 x= x +v t + at = at (2) At point ,B x=xB =823 m and t =30 s
(a) Solving (2) for a,
( )( )
( )
2 2 2 823 2 30 x a t = = m/s2 a= t (b) Then, vB =at =( )( )
1.8 30 vB =54 m/s t 1.8 Constant acceleration: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.
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 v and travels 0 165 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 Constant acceleration. x0 =0 0 v=v +at (1) 2 0 0 1 2 x= x +v t + at (2) Solving (1) for a, 0 v v a t − = (3) Then, 0 0 1 0 2 0 1
(
0)
1(
0)
2 2 2 v v x x v t t x v v t v v t t − = + + = + + = + At t =6 s, 1 0 and 6 165 m 2 v= v x =( )
0 0 0 0 0 1 1 165 165 6 4.5 or 36.7 m/s 2 2 4.5 1 18.3 m/s 2 v v v v v v = + = = = = = Then, from (3), 18.3 36.7 m/s2 3.05 m/s2 6 6 a= − = − = −Substituting into (1) and (2), v=36.7 3.05− t
( )
2 1 0 2 x= + t− t At stopping, v=0 or − ts =0 ts =12 s(
)( )
1(
)( )
2 0 12 12 220.8 m 2 x= + − =( ) Additional time for stoppinga =12 s−6 s ∆ =t 6 s t
( ) Additional distance for stoppingb =220.8 m 165 m− ∆ =d 55.8 m t
18.3 36.7 3.05 36.7 3.05 36.7 3.05 229-m Constant acceleration: 0,
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.
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
2 0 0
1 ( ) During the acceleration phase
2
a x=x +v t + at
0 0
Using x =0, and v =0, and solving for givesa
2
2x a
t = Noting that x=130 m when t =25 s,
( )( )
( )
22 130 25
a= a=0.416 m/s t
(b) Final velocity is reached at t =25 s.
(
)( )
0 0 0.416 25
f
v =v +at= + vf =10.40 m/s t
(c) The remaining distance for the constant speed phase is
400 130 270 m
x
∆ = − =
For constant velocity, 270 25.96 s
10.40 x t v ∆ ∆ = = =
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.
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
v
1 of the rocket at the end of powered flight, (b) the maximum altitude reached by the rocket.SOLUTION
Constant acceleration. Choose t = at end of powered flight. 0
Then, y1 =27.5 m a= − = −g 9.81 m/s2
(a) When y reaches the ground, yf =0 and t =16 s.
2 2 1 1 1 1 1 1 2 2 f y = y +v t+ at = y +v t− gt
( )( )
2 2 1 1 1 2 2 1 0 27.5 9.81 16 76.76 m/s 16 f y y gt v t − + − + = = = 1 76.8 m/s v = t(b) When the rocket reaches its maximum altitude ymax, 0 v=
(
)
(
)
2 2 2 1 2 1 1 2 1 v =v + a y− y =v − g y − y 2 2 1 1 2 v v y y g − = −(
)
( )( )
2 max 0 76.76 27.5 2 9.81 y = − − ymax =328 m t Constant acceleration: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.
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 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. Motion of auto.
( )
( )
2 0 0, 0 0, 0.75 m/s A A A x = v = a =( ) ( )
2(
)
2 0 0 1 1 0 0 0.75 2 2 A A A A x = x + v t + a t = + + t 2 0.375 m A x = t Motion of bus.( )
xB 0=?,( )
vB 0 = −6 m/s, aB =0( ) ( )
0 0( )
0 6 m B B B B x = x − v t= x − t At t =20 ,s xB =0.( ) ( )( )
0 0= xB − 6 20( )
0 120 m B x = Hence, xB =120−6tWhen the vehicles pass each other, xB = xA.
2 120−6t =0.375t 2 0.375t +6t −120=0
( )(
)(
)
( )(
)
2 6 (6) 4 0.375 120 2 0.375 t = − ± − − 6 14.697 11.596 s and 27.6 s 0.75 t = − ± = −Reject the negative root. t =11.60 st
Corresponding values of xAand xB.
(
)(
)
2 0.375 11.596 50.4 m A x = =( )(
)
120 6 11.596 50.4 m B x = − = x=50.4 mt at 0: ,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.
Automobiles A and B are traveling in adjacent highway lanes and at 0
t = 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:
( )
( )
2 0 0, 0 15 km/h = 4.1667 m/s, 0.6 m/s A A A x = v = a =( )
0 4.1667 0.6 A A A v = v +a t = + t( ) ( )
2 2 0 0 1 4.1667 0.3 2 A A A A x = x + v t + a t = t + t Motion of B:( )
( )
2 0 25 m, 0 23 km/h = 6.3889 m/s, 0.4 m/s B B B x = v = a = −( )
0 6.3889 0.4 B B B v = v +a t = − t( ) ( )
2 2 0 0 1 25 6.3889 0.2 2 B B B B x = x + v t + a t = + t − t(a) When and where A overtakes B. xA =xB
2 2 4.1667t +0.3t =25+6.3889t −0.2t 2 0.5t −2.2222t −25=0
( )( )( )
( )( )
2 2.2222 2.2222 4 0.5 25 2 0.5 t = ± − − 2.2222 7.4120 9.6343 s and 5.19 s t = ± = −Reject the negative root. .t =9.63 st
(
)(
) ( )(
)
2 4.1667 9.6343 0.3 9.6343 68.0 m A x = + =(
)(
) ( )(
)
2 25 6.3889 9.6343 0.2 9.6343 68.0 m B x = + − = moves 68.0 m A t moves 43.0 m B t (b) Corresponding speeds.( )(
)
4.1667 0.6 9.6343 9.947 m/s A v = + = vA=35.8 km/ht( )(
)
6.3889 0.4 9.6343 2.535 m/s B v = − = vB =9.13 km/ht B: speeds: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.
49
PROBLEM 11.43
In a close harness race, horse 2 passes horse 1 at point A, where the two velocities are
v
2 = 6.4 m/s andv
1 = 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 aret
1 = 61.5 s and2
t
= 62.0 s, respectively. Assuming uniform accelerations for both horses betweenA 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. Let x= and 0 t = when the horses are at point A.0Then, 0 1 2 2 x = v t + at Solving for , a
(
0)
2 2 x v t a t − =Using x=366 m and the initial velocities and elapsed times for each horse,
( )(
)
(
)
–3 1 1 1 2 2 1 2 366 6.2 61.5 8× 10 m/s 61.5 x v t a t − − = = = −( )(
)
(
)
2 2 2 2 2 2 2 2 6.4 62.0 1.6 10 m/s 62.0 x v t a t − − = = = − 1 2 Calculating x −x , 1 2(
1 2)
1(
1 2)
2 2 x −x = v −v t + a −a t(
)
(
) (
)
2 1 2 2 1 2 .2 .004 x x t t t t − = − + − − − = − + At point B, x1− x2 =0 − tB + tB2 =0 (a) 0.2 50 s .004 B t = =Calculating x using data for either horse, B
Horse 1:
( )( )
1(
)( )
2 2 B x = + − xB =340 m� Horse 2:( )( )
1(
)( )
2 340 m 2 B x = + − =When horse 1 crosses the finish line at t =61.5 s,
(b) x1 −x2 = −
( )(
61.5) ( )(
+ 61.5)
2 ∆ =x 2.8 m � − 2 366 − 2 × 6.2 6.4 8 × 10−–3 1.6 10 −× 2 .004 50 6.2 8 × 10− 50–3 50 6.4 1.6 10 −× 2 50 0.2 .004 0.2 2: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.
Two rockets are launched at a fireworks performance. Rocket A is launched with an initial velocity v and rocket B is launched 4 s later 0 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 v (b) the velocity of B relative to A at the time of the 0, explosion.
SOLUTION
Choose x positive upward. Constant acceleration a= −g
Rocket launch data: Rocket :A x=0, v=v0, t = 0
Rocket : B x=0, v=v0, t =tB =4 s Velocities: Rocket :A vA=v0− gt Rocket : B vB =v0 −g t
(
−tB)
Positions: Rocket : 0 1 2 2 A A x =v t − gt(
)
(
)
2 0 1 Rocket : , 2 B B B B B x =v t −t − g t −t t ≥tFor simultaneous explosions at xA =xB =73 m when t =tE,
(
)
(
)
2 2 2 2 0 0 0 0 1 1 1 1 2 2 2 2 E E E B E B E B E E B B v t − gt =v t −t − g t −t =v t −v t − gt + gt t − gt 0 Solving for , v 0 2 B E gt v =gt − (1) Then, when t =tE, 1 2, 2 2 B A E E E gt x =gt − t − gt or 2 2 A 0 E B E x t t t g − − = Solving for , tE( )( )
( )
( )
( )( )( )( ) 2 2 2 4 1 2 73 9.81 4 1 4 4 6.35 s 2 2 A x B B g E t t t ± + ± + = = =Assuming a constant acceleration g = 9.81 m/s2, determine (a) the initial
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.
51
PROBLEM 11.44 CONTINUED
(a) From equation (1), 0
( )(
9.81 6.35) (
9.81 4)( )
2 v = − v0=42.67 m/s � At time ,tE vA =v0−gtE vB =v0 − g t(
E −tB)
(b) vB−vA =gtB =( )( )
9.81 4 vB A/ =39.24 m/s � From Equation (1), ,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.
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 = , the boats accelerate 0 at constant rates. Knowing that when B passes A, t = s and 8
228 A
v = km/h, determine (a) the acceleration of A, (b) the acceleration ofB. SOLUTION (a) Acceleration of A.
