Covers Chapters 12 to 22
SOLUTIONS & ANSWERS
,TABLE OF CONTENTS
12. Kinematics of a Particle
13. Kinetics of a Particle: Force and Acceleration
14. Kinetics of a Particle: Work and Energy
15. Kinetics of a Particle: Impulse and Momentum
16. Planar Kinematics of a Rigid Body
17. Planar Kinetics of a Rigid Body: Force and Acceleration
18. Planar Kinetics of a Rigid Body: Work and Energy
19. Planar Kinetics of a Rigid Body: Impulse and Momentum
20. Three-Dimensional Kinematics of a Rigid Body
21. Three-Dimensional Kinetics of a Rigid Body
22. Vibrations
,© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–1.
Starting from rest, a particle moving in a straight line has an
acceleration of a = (2t - 6) m>s2, where t is in seconds. What
is the particle’s velocity when t = 6 s, and what is its position
when t = 11 s?
SOLUTION
a = 2t - 6
dv = a dt
v t
L0 L0
dv = (2t - 6) dt
v = t 2 - 6t
ds = v dt
s t
L0 L0
ds = (t2 - 6t) dt
t3
s = - 3t2
3
When t = 6 s,
v = 0 Ans.
When t = 11 s,
s = 80.7 m Ans.
Ans:
s = 80.7 m
1
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–2.
If a particle has an initial velocity of v0 = 12 ft>s to the
right, at s0 = 0, determine its position when t = 10 s, if
a = 2 ft>s2 to the left.
SOLUTION
1 2
1S
+2 s = s0 + v0 t + a t
2 c
1
= 0 + 12(10) + ( -2)(10)2
2
= 20 ft Ans.
Ans:
s = 20 ft
2
,© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–3.
A particle travels along a straight line with a velocity
v = (12 - 3t 2) m>s, where t is in seconds. When t = 1 s, the
particle is located 10 m to the left of the origin. Determine
the acceleration when t = 4 s, the displacement from
t = 0 to t = 10 s, and the distance the particle travels during
this time period.
SOLUTION
v = 12 - 3t 2 (1)
dv
a = = - 6t t=4 = -24 m>s2 Ans.
dt
s t t
L-10 L1 L1
ds = v dt = ( 12 - 3t 2 ) dt
s + 10 = 12t - t 3 - 11
s = 12t - t 3 - 21
s t=0 = - 21
s t = 10 = -901
∆s = -901 - ( -21) = -880 m Ans.
From Eq. (1):
v = 0 when t = 2s
s t=2 = 12(2) - (2)3 - 21 = -5
sT = (21 - 5) + (901 - 5) = 912 m Ans.
Ans:
a = - 24 m>s2
∆s = -880 m
sT = 912 m
3
,© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–4.
A particle travels along a straight line with a constant
acceleration. When s = 4 ft, v = 3 ft>s and when s = 10 ft,
v = 8 ft>s. Determine the velocity as a function of position.
SOLUTION
Velocity: To determine the constant acceleration ac, set s0 = 4 ft, v0 = 3 ft>s,
s = 10 ft and v = 8 ft>s and apply Eq. 12–6.
+ )
(: v2 = v20 + 2ac (s - s0)
82 = 32 + 2ac (10 - 4)
ac = 4.583 ft>s2
Using the result ac = 4.583 ft>s2, the velocity function can be obtained by applying
Eq. 12–6.
+ )
(: v2 = v20 + 2ac (s - s0)
v2 = 32 + 2(4.583) (s - 4)
A
v = A 29.17s - 27.7 ft>s Ans.
v = ( 19.17s - 27.7 ) ft>s
Ans:
4
,© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–5.
The velocity of a particle traveling in a straight line is given
by v = (6t - 3t2) m>s, where t is in seconds. If s = 0 when
t = 0, determine the particle’s deceleration and position
when t = 3 s. How far has the particle traveled during the
3-s time interval, and what is its average speed?
SOLUTION
v = 6t - 3t 2
dv
a = = 6 - 6t
dt
At t = 3 s
a = - 12 m>s2 Ans.
ds = v dt
s t
L0 L0
ds = (6t - 3t2)dt
s = 3t 2 - t 3
At t = 3 s
s = 0 Ans.
Since v = 0 = 6t - 3t 2, when t = 0 and t = 2 s.
when t = 2 s, s = 3(2)2 - (2)3 = 4 m
sT = 4 + 4 = 8 m Ans.
sT 8
( vsp ) avg = = = 2.67 m>s Ans.
t 3
Ans:
sT = 8 m
vavg = 2.67 m>s
5
,© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–6.
The position of a particle along a straight line is given by
s = (1.5t3 - 13.5t2 + 22.5t) ft, where t is in seconds.
Determine the position of the particle when t = 6 s and the
total distance it travels during the 6-s time interval. Hint:
Plot the path to determine the total distance traveled.
SOLUTION
Position: The position of the particle when t = 6 s is
s|t = 6s = 1.5(63) - 13.5(62) + 22.5(6) = -27.0 ft Ans.
Total DistanceTraveled: The velocity of the particle can be determined by applying
Eq. 12–1.
ds
v = = 4.50t2 - 27.0t + 22.5
dt
The times when the particle stops are
4.50t2 - 27.0t + 22.5 = 0
t = 1s and t = 5s
The position of the particle at t = 0 s, 1 s and 5 s are
s t = 0 s = 1.5(03) - 13.5(02) + 22.5(0) = 0
s t = 1 s = 1.5(13) - 13.5(12) + 22.5(1) = 10.5 ft
s t = 5 s = 1.5(53) - 13.5(52) + 22.5(5) = -37.5 ft
From the particle’s path, the total distance is
stot = 10.5 + 48.0 + 10.5 = 69.0 ft Ans.
Ans:
s
t = 6 s = - 27.0 ft
stot = 69.0 ft
6
,© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–7.
A particle moves along a straight line such that its position
is defined by s = (t2 - 6t + 5) m. Determine the average
velocity, the average speed, and the acceleration of the
particle when t = 6 s.
SOLUTION
s = t2 - 6t + 5
ds
v = = 2t - 6
dt
dv
a = = 2
dt
v = 0 when t = 3
s
t=0 = 5
s
t = 3 = -4
s
t=6 = 5
∆s 0
vavg = = = 0 Ans.
∆t 6
sT 9 + 9
( vsp ) avg = = = 3 m>s Ans.
∆t 6
a
t = 6 = 2 m>s2 Ans.
Ans:
vavg = 0
(vsp)avg = 3 m>s
a
t = 6 s = 2 m>s2
7
, © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–8.
A particle is moving along a straight line such that its
position is defined by s = (10t2 + 20) mm, where t is in
seconds. Determine (a) the displacement of the particle
during the time interval from t = 1 s to t = 5 s, (b) the
average velocity of the particle during this time interval,
and (c) the acceleration when t = 1 s.
SOLUTION
s = 10t2 + 20
(a) s|1 s = 10(1)2 + 20 = 30 mm
s|5 s = 10(5)2 + 20 = 270 mm
¢s = 270 - 30 = 240 mm Ans.
(b) ¢t = 5 - 1 = 4 s
¢s 240
vavg = = = 60 mm>s Ans.
¢t 4
d2s
(c) a = = 20 mm s2 (for all t) Ans.
dt2
Ans:
∆s = 240 mm
vavg = 60 mm>s
a = 20 mm>s2
8
SOLUTIONS & ANSWERS
,TABLE OF CONTENTS
12. Kinematics of a Particle
13. Kinetics of a Particle: Force and Acceleration
14. Kinetics of a Particle: Work and Energy
15. Kinetics of a Particle: Impulse and Momentum
16. Planar Kinematics of a Rigid Body
17. Planar Kinetics of a Rigid Body: Force and Acceleration
18. Planar Kinetics of a Rigid Body: Work and Energy
19. Planar Kinetics of a Rigid Body: Impulse and Momentum
20. Three-Dimensional Kinematics of a Rigid Body
21. Three-Dimensional Kinetics of a Rigid Body
22. Vibrations
,© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–1.
Starting from rest, a particle moving in a straight line has an
acceleration of a = (2t - 6) m>s2, where t is in seconds. What
is the particle’s velocity when t = 6 s, and what is its position
when t = 11 s?
SOLUTION
a = 2t - 6
dv = a dt
v t
L0 L0
dv = (2t - 6) dt
v = t 2 - 6t
ds = v dt
s t
L0 L0
ds = (t2 - 6t) dt
t3
s = - 3t2
3
When t = 6 s,
v = 0 Ans.
When t = 11 s,
s = 80.7 m Ans.
Ans:
s = 80.7 m
1
,© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–2.
If a particle has an initial velocity of v0 = 12 ft>s to the
right, at s0 = 0, determine its position when t = 10 s, if
a = 2 ft>s2 to the left.
SOLUTION
1 2
1S
+2 s = s0 + v0 t + a t
2 c
1
= 0 + 12(10) + ( -2)(10)2
2
= 20 ft Ans.
Ans:
s = 20 ft
2
,© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–3.
A particle travels along a straight line with a velocity
v = (12 - 3t 2) m>s, where t is in seconds. When t = 1 s, the
particle is located 10 m to the left of the origin. Determine
the acceleration when t = 4 s, the displacement from
t = 0 to t = 10 s, and the distance the particle travels during
this time period.
SOLUTION
v = 12 - 3t 2 (1)
dv
a = = - 6t t=4 = -24 m>s2 Ans.
dt
s t t
L-10 L1 L1
ds = v dt = ( 12 - 3t 2 ) dt
s + 10 = 12t - t 3 - 11
s = 12t - t 3 - 21
s t=0 = - 21
s t = 10 = -901
∆s = -901 - ( -21) = -880 m Ans.
From Eq. (1):
v = 0 when t = 2s
s t=2 = 12(2) - (2)3 - 21 = -5
sT = (21 - 5) + (901 - 5) = 912 m Ans.
Ans:
a = - 24 m>s2
∆s = -880 m
sT = 912 m
3
,© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–4.
A particle travels along a straight line with a constant
acceleration. When s = 4 ft, v = 3 ft>s and when s = 10 ft,
v = 8 ft>s. Determine the velocity as a function of position.
SOLUTION
Velocity: To determine the constant acceleration ac, set s0 = 4 ft, v0 = 3 ft>s,
s = 10 ft and v = 8 ft>s and apply Eq. 12–6.
+ )
(: v2 = v20 + 2ac (s - s0)
82 = 32 + 2ac (10 - 4)
ac = 4.583 ft>s2
Using the result ac = 4.583 ft>s2, the velocity function can be obtained by applying
Eq. 12–6.
+ )
(: v2 = v20 + 2ac (s - s0)
v2 = 32 + 2(4.583) (s - 4)
A
v = A 29.17s - 27.7 ft>s Ans.
v = ( 19.17s - 27.7 ) ft>s
Ans:
4
,© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–5.
The velocity of a particle traveling in a straight line is given
by v = (6t - 3t2) m>s, where t is in seconds. If s = 0 when
t = 0, determine the particle’s deceleration and position
when t = 3 s. How far has the particle traveled during the
3-s time interval, and what is its average speed?
SOLUTION
v = 6t - 3t 2
dv
a = = 6 - 6t
dt
At t = 3 s
a = - 12 m>s2 Ans.
ds = v dt
s t
L0 L0
ds = (6t - 3t2)dt
s = 3t 2 - t 3
At t = 3 s
s = 0 Ans.
Since v = 0 = 6t - 3t 2, when t = 0 and t = 2 s.
when t = 2 s, s = 3(2)2 - (2)3 = 4 m
sT = 4 + 4 = 8 m Ans.
sT 8
( vsp ) avg = = = 2.67 m>s Ans.
t 3
Ans:
sT = 8 m
vavg = 2.67 m>s
5
,© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–6.
The position of a particle along a straight line is given by
s = (1.5t3 - 13.5t2 + 22.5t) ft, where t is in seconds.
Determine the position of the particle when t = 6 s and the
total distance it travels during the 6-s time interval. Hint:
Plot the path to determine the total distance traveled.
SOLUTION
Position: The position of the particle when t = 6 s is
s|t = 6s = 1.5(63) - 13.5(62) + 22.5(6) = -27.0 ft Ans.
Total DistanceTraveled: The velocity of the particle can be determined by applying
Eq. 12–1.
ds
v = = 4.50t2 - 27.0t + 22.5
dt
The times when the particle stops are
4.50t2 - 27.0t + 22.5 = 0
t = 1s and t = 5s
The position of the particle at t = 0 s, 1 s and 5 s are
s t = 0 s = 1.5(03) - 13.5(02) + 22.5(0) = 0
s t = 1 s = 1.5(13) - 13.5(12) + 22.5(1) = 10.5 ft
s t = 5 s = 1.5(53) - 13.5(52) + 22.5(5) = -37.5 ft
From the particle’s path, the total distance is
stot = 10.5 + 48.0 + 10.5 = 69.0 ft Ans.
Ans:
s
t = 6 s = - 27.0 ft
stot = 69.0 ft
6
,© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–7.
A particle moves along a straight line such that its position
is defined by s = (t2 - 6t + 5) m. Determine the average
velocity, the average speed, and the acceleration of the
particle when t = 6 s.
SOLUTION
s = t2 - 6t + 5
ds
v = = 2t - 6
dt
dv
a = = 2
dt
v = 0 when t = 3
s
t=0 = 5
s
t = 3 = -4
s
t=6 = 5
∆s 0
vavg = = = 0 Ans.
∆t 6
sT 9 + 9
( vsp ) avg = = = 3 m>s Ans.
∆t 6
a
t = 6 = 2 m>s2 Ans.
Ans:
vavg = 0
(vsp)avg = 3 m>s
a
t = 6 s = 2 m>s2
7
, © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–8.
A particle is moving along a straight line such that its
position is defined by s = (10t2 + 20) mm, where t is in
seconds. Determine (a) the displacement of the particle
during the time interval from t = 1 s to t = 5 s, (b) the
average velocity of the particle during this time interval,
and (c) the acceleration when t = 1 s.
SOLUTION
s = 10t2 + 20
(a) s|1 s = 10(1)2 + 20 = 30 mm
s|5 s = 10(5)2 + 20 = 270 mm
¢s = 270 - 30 = 240 mm Ans.
(b) ¢t = 5 - 1 = 4 s
¢s 240
vavg = = = 60 mm>s Ans.
¢t 4
d2s
(c) a = = 20 mm s2 (for all t) Ans.
dt2
Ans:
∆s = 240 mm
vavg = 60 mm>s
a = 20 mm>s2
8
