PROBLEM 11.

A loaded railroad car is rolling at a constant velocity

when it couples with a spring and dashpot bumper
system. After the coupling, the motion of the car is
4.8t
defined by the relation x  60e sin16t where x and t
are expressed in mm and seconds, respectively.
Determine the position, the velocity and the
acceleration of the railroad car when (a) ( t 0,
(b) t 0.3 s.

SOLUTION

x  60e4.8t sin16t
dx
v  60(4.8)e 4.8t sin16t  60(16)e 4.8t cos16t
dt
v  288e 4.8t sin16t  960e4.8t cos16t
dv
a  1382.4e 4.8t sin16t  4608e 4.8t cos16t
dt
 4608e4.8t cos16t  15360e 4.8t sin16t
a  13977.6e 4.8t sin16t  9216e 4.8 cos16t
(a) At t  0, x0  0 x0  0 mm 

v0  960 mm/s v0  960 mm/s 

a0  9216 mm/s 2 a0  9220 mm/s 2 

(b) At t  0.3 s, e 4.8t  e1.44  0.23692

sin16t  sin 4.8  0.99616
cos16t  cos 4.8  0.08750
x0.3  (60)(0.23692)(0.99616)  14.16 x0.3  14.16 mm 

v0.3  (288)(0.23692)(0.99616)
 (960)(0.23692)(0.08750)  87.9 v0.3  87.9 mm/s 

a0.3  (13977.6)(0.23692)( 0.99616)

 (9216)(0.23692)(0.08750)  3108 a0.3  3110 mm/s 2 
2
or 3.11 m/s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

6
PROBLEM 11.5

The motion of a particle is defined by the relation x  6t 4  2t 3  12t 2  3t  3, where x and t are expressed in
meters and seconds, respectively. Determine the time, the position, and the velocity when a  0.

SOLUTION

We have x  6t 4  2t 3  12t 2  3t  3
dx
Then v  24t 3  6t 2  24t  3
dt
dv
and a  72t 2  12t  24
dt

When a  0: 72t 2  12t  24  12(6t 2  t  2)  0

or (3t  2)(2t  1)  0

2 1
or t s and t   s (Reject) t  0.667 s 
3 2
4 3 2
2 2 2 2 2
At t  s: x2/3  6 2  12 3 3 or x2/3  0.259 m 
3 3 3 3 3
3 2
2 2 2
v2/3  24 6  24 3 or v2/3  8.56 m/s 
3 3 3

McGraw-Hill Education.
Copyright © McGraw-Hill Education. Permission
Permission required
required for
forreproduction
reproductionor
ordisplay.
display.
7
 PROBLEM 11.9

The brakes of a car are applied, causing it to slow down at a

rate of 3 m/s2. Knowing that the car stops in 100 m, determine
(a) how fast the car was traveling immediately before the
brakes were applied, (b) the time required for the car to stop.

SOLUTION

a = −3 m/s 2
(a) Velocity at x = 0.
dv
v = a = −3
dx
0 x
∫ v0 vdv = − ∫ 0 (−3)dx
f

v02
0− = −3x f = −(3)(100)
2
v02 = 600 v0 = 24.5 m/s 2
(b) Time to stop.
dv
= a = −3
dx
0 tf
∫ v0 dv = − ∫ 0 −3dt
0 − v0 = −3t f
v0 24.5
tf = = t f = 8.17 s
3 3

PROPRIETARY MATERIAL. © 2013 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.

Copyright © McGraw-Hill Education. Permission

13 required for reproduction or display.
11
 PROBLEM 11.16

A projectile enters a resisting medium at x  0 with an initial velocity

v0  900 ft/s and travels 4 in. before coming to rest. Assuming that the
velocity of the projectile is defined by the relation v  v0  kx, where v is
expressed in ft/s and x is in feet, determine (a) the initial acceleration of the
projectile, (b) the time required for the projectile to penetrate 3.9 in. into the
resisting medium.

SOLUTION
First note
4 4 
When x  ft, v  0: 0  (900 ft/s)  k  ft 
12  12 
1
or k  2700
s
(a) We have v  v0  kx
dv d
Then a  (v0  kx)  kv
dt dt
or a  k (v0  kx)
1
At t  0: a  2700 (900 ft/s  0)
s

or a0  2.43  106 ft/s 2 

dx
(b) We have  v  v0  kx
dt
x dx t
At t  0, x  0: 0 v0  kx
  dt
0

1
or  [ln(v0  kx)]0x  t
k

1  v0  1  1 
or t ln    ln  
k  v0  kx  k  1  vk x 
 0 

1  1 
When x  3.9 in.: t ln  2700 1/s 3.9 
900 ft/s  12 
1
2700 s  1  ft

or t  1.366  103 s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

18
 PROBLEM 11.41

As relay runner A enters the 20-m-long exchange zone with a

speed of 12.9 m/s, he begins to slow down. He hands the baton to
runner B 1.82 s later as they leave the exchange zone with the
same velocity. Determine (a) the uniform acceleration of each of
the runners, (b) when runner B should begin to run.

SOLUTION

1
(a) For runner A: x A = 0 + (v A ) 0 t + a At 2
2
1
At t = 1.82 s: 20 m = (12.9 m/s)(1.82 s) + a A (1.82 s) 2
2

or a A = −2.10 m/s2

Also vA = (vA )0 + a At

At t = 1.82 s: (v A )1.82 = (12.9 m/s) + (−2.10 m/s 2 )(1.82 s)

= 9.078 m/s

For runner B: vB2 = 0 + 2aB [ xB − 0 ]

When xB = 20 m, vB = v A : (9.078 m/s)2 = 2aB (20 m)

or aB = 2.0603 m/s 2

aB = 2.06 m/s2

(b) For runner B: vB = 0 + aB (t − t B )

where t B is the time at which he begins to run.

At t = 1.82 s: 9.078 m/s = (2.0603 m/s2 )(1.82 − t B )s

or t B = −2.59 s
Runner B should start to run 2.59 s before A reaches the exchange zone.

PROPRIETARY MATERIAL. © 2013 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.

Copyright © McGraw-Hill Education. Permission

48 required for reproduction or display.
45
 PROBLEM 11.53

Slider block A moves to the left with a constant velocity of 6 m/s.

Determine (a) the velocity of block B, (b) the velocity of portion D of
the cable, (c) the relative velocity of portion C of the cable with respect
to portion D.

SOLUTION

From the diagram, we have

x A + 3 yB = constant

Then v A + 3vB = 0 (1)

and a A + 3aB = 0 (2)

(a) Substituting into Eq. (1) 6 m/s + 3vB = 0

or v B = 2.00 m/s �

(b) From the diagram yB + yD = constant

Then vB + vD = 0

v D = 2.00 m/s �

(c) From the diagram x A + yC = constant

Then v A + vC = 0 vC = −6 m/s

Now vC/D = vC − vD = (−6 m/s) − (2 m/s) = −8 m/s

v C/D = 8.00 m/s �

PROPRIETARY MATERIAL. © 2013 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.

Copyright © McGraw-Hill Education. Permission

61 required for reproduction or display.
62
 PROBLEM 11.98

A ski jumper starts with a horizontal take-off velocity of 25

m/s and lands on a straight landing hill inclined at 30o.
Determine (a) the time between take-off and landing, (b) the
length d of the jump, (c) the maximum vertical distance
between the jumper and the landing hill.

SOLUTION

(a) At the landing point, y   x tan 30

Horizontal motion: x  x0   v x 0 t  v0t

Vertical motion:  0
y  y0  v y t 
1 2
2
1
gt   gt 2
2

2y 2 x tan 30 2v0t tan 30

from which t2    
g g g

2v0 tan 30  2  25  tan 30

Rejecting the t  0 solution gives t   t  2.94 s 
g 9.81

(b) Landing distance: d 

x

v0t

 25 2.94  d  84.9 m 
cos 30 cos 30 cos 30

(c) Vertical distance: h  x tan 30  y

1 2
or h  v0t tan 30  gt
2

Differentiating and setting equal to zero,

dh vo tan 30
 v0 tan 30  gt  0 or t 
dt g

 v0  v0 tan 30 tan 30  1 g  v0 tan 30 

2

Then, hmax   
g 2  g 

v 2 tan 2 30  25  tan 30 

2 2
 0  hmax  10.62 m 
2g  2 9.81

Copyright ©
Copyright © McGraw-Hill
McGraw-Hill Education.
Education.Permission
Permissionrequired
requiredfor
forreproduction
reproductionorordisplay.
display.
132
PROBLEM 11.108

A tennis player serves the ball at a height h 2.5 m with an initial velocity of v0 at an angle of 5° with the
horizontal. Determine the range for which of v0 for which the ball will land in the service area which extends
to 6.4 m beyond the net.

SOLUTION
The motion is projectile motion. Place the origin of the xy-coordinate system at ground level just below the
point where the racket impacts the ball. The coordinates of this impact point are x0  0, y0  h  2.5 m. The
components of initial velocity are (vx )0  v0 cos5 and (v y )0  v0 sin 5.

Horizontal motion: x  x0  (vx )0 t  (v0 cos5)t (1)

1 2
Vertical motion: y  y0  (v y )0 t  gt
2
1
 2.5  (v0 sin 5)t   (9.81)t 2 (2)
2
x
From (1), v0t  (3)
cos5

Then y  2.5  x tan 5  4.905t 2

2.5  x tan 5  y
t2  (4)
4.905
At the minimum speed the ball just clears the net.
x  12.2 m, y  0.914 m
12.2
v0 t   12.2466 m
cos 5
2.5  12.2 tan 5  0.914
t2  t  0.32517 s
4.905
12.2466
v0  v0  37.66 m/s
0.32517

McGraw-Hill Education.
Copyright © McGraw-Hill Education. Permission
Permission required
required for
forreproduction
reproductionor
ordisplay.
display.
145
 PROBLEM 11.108 (Continued)

At the maximum speed the ball lands 6.4 m beyond the net.
x  12.2  6.4  18.6 m y0
18.6
v0t   18.6710 m
cos 5
2.5  18.6 tan 5  0
t2  t  0.42181 s
4.905
18.6710
v0  v0  44.26 m/s
0.42181
Range for v0. 37.7 m/s  v0  44.3 m/s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

146
 PROBLEM 11.123

Knowing that the velocity of block B with respect to block A is

vB/A = 5.6 m /s 70°, determine the velocities of A and B.

SOLUTION

From the diagram

2 x A + 3x B = constant

Then 2v A + 3v B = 0
2
or | vB | = v A
3
Now v B = v A + v B /A

and noting that vA and vB must be parallel to surfaces A and B, respectively, the graphical representation of this
equation is then as shown. Note: Assuming that vA is directed up the incline leads to a velocity diagram that
does not “close.”

First note a = 180∞ - ( 40∞ + 30∞ + q B )

= 110∞ - q B
2
vA vA 5.6
Then = 3 =
sin (110∞ - q B ) sin 40∞ sin (30∞ + q B )
2
or v A sin 40∞ = v A sin (110∞ - q B )
3
or sin (110∞ - q B ) = 0.96418

or q B = 35.3817∞

and q B = 4.6183∞
2 5.6 sin 40∞
For q B = 35.3817∞ : vB = vA =
3 sin (30∞ + 35.3817∞)
or v A = 5.94 m/s v B = 3.96 m/s

v A = 5.94 m/s 30° !

v B = 3.96 m/s 35.4° !

PROPRIETARY MATERIAL. © 2010 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.

Copyright © McGraw-Hill Education. Permission

180 required for reproduction or display.
170
 PROBLEM 11.123 (Continued)

2 5.6 sin 40∞

For q B = 4.6183∞ : vB = vA =
3 sin (30∞ + 4.6183∞)
or v A = 9.50 m/s

v B = 6.34 m/s
v A = 9.50 m/s 30° !

v B = 6.34 m/s 4.62° !

PROPRIETARY MATERIAL. © 2010 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.

Copyright © McGraw-Hill Education. Permission

181 required for reproduction or display.
171
 PROBLEM 11.143
A race car enters the circular portion of a track that has a radius
of 70 m. When the car enters the curve at point P, it is
travelling with a speed of 120 km/h that is increasing at 5 m/s2.
Three seconds later, determine (a) the total acceleration of the
car in xy components, (b) the linear velocity of the car in xy
components.

SOLUTION

Speeds: vto  120 km/h  33.33 m/s

vt  vto  at t
vt  33.33  5(3)  48.33 m/s
1
Distance: s  so  vto t  at t 2
2
1
s  33.33(3)  (5)(3)2  122.5 m
2
Location: sR
s 122.5
   1.75 rad
R 70
 100.3
Tangential Acceleration: at  5 m/s 2
v2
Normal Acceleration: an 

48.332
an   33.37 m/s 2
70
(a) Accelerations: ax  5cos(10.3)  33.37sin(10.3)
ax  1.047 m/s 2 

a y  5sin(10.3)  33.37 cos(10.3)

a y  33.726 m/s 2 

(b) Velocities: vx  48.33cos(10.3)

vx  47.55 m/s 
v y  48.33sin(10.3)
v y  8.64 m/s 

McGraw-Hill Education.
Copyright © McGraw
McGraw-Hill Education. Permission
Permission required
required for
forreproduction
reproductionor
ordisplay.
display.
195
 PROBLEM 11.163

During a parasailing ride, the boat is traveling at

a constant 30 km/hr with a 200 m long tow line.
At the instant shown, the angle between the line
and the water is 30º and is increasing at a
constant rate of 2º/s. Determine the velocity and
acceleration of the parasailer at this instant.

SOLUTION

Given: v B  30i km/hr  8.333i m/s

aB  0 ê
r  200 m, r  0, r  0 eˆr
r
  30 P
  2 / s  0.0349 rad/s 
=0 B

Relative Motion relations: vP  vB  vP/ B

aP  aB  aP / B
Using Radial and Transverse components:
v P / B  rer  re

a P / B     
r  r2 er  r  2r e
Substitute in known values: v P / B  6.981e m/s
a P / B  0.2437er m/s 2
Change to rectangular coordinates:
v P / B  6.981*sin 30i  6.981*cos 30 j m/s
=3.491i  6.046 j m/s
a P / B  0.2437*   cos 30  i  0.2437 *sin 30 j m/s 2
 0.2111i  0.1219 j m/s 2
Substitute into Relative Motion relations:
v P  8.33i  3.491i  6.046 j m/s
=11.824i  6.046 j m/s
a P  0  0.2111i  0.1219 j m/s 2
 0.2111i  0.1219 j m/s 2
Velocity: vP  13.280 m/s 27.08 
Acceleration: aP  0.2437 m/s 2
30.00 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

218