(Engineering Mechanics_ Dynamics 15th Edition Russell C:Q&A).
Engineering Mechanics: Dynamics 15th Edition Russell C. Hibbeler – Comprehensive
Study Material and Practice Problems
Page 1 of 238
, (Engineering Mechanics_ Dynamics 15th Edition Russell C:Q&A).
This document covers fundamental concepts in engineering mechanics with a focus on
dynamics, including kinematics, kinetics, particle motion, rigid body motion, and energy
methods. It includes practice problems with solutions to support understanding and exam
preparation. The material is aligned with the 15th edition by Russell C. Hibbeler and is ideal
for engineering students studying dynamics and mechanical systems.
Page 2 of 238
, (Hibbeler 15th dynamics solution manual:Q&A).
© 2022 by R.C. Hibbeler. Published by Pearson Education, Inc., Hoboken, 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:
v = 0
s = 80.7 m
1
Page 3 of 238
, (Hibbeler 15th dynamics solution manual:Q&A).
© 2022 by R.C. Hibbeler. Published by Pearson Education, Inc., Hoboken, 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
Page 4 of 238
, (Hibbeler 15th dynamics solution manual:Q&A).
© 2022 by R.C. Hibbeler. Published by Pearson Education, Inc., Hoboken, 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
2
v = (12 -
v - 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 t == 44 s,s, the
acceleration when the displacement
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
Page 5 of 238
Engineering Mechanics: Dynamics 15th Edition Russell C. Hibbeler – Comprehensive
Study Material and Practice Problems
Page 1 of 238
, (Engineering Mechanics_ Dynamics 15th Edition Russell C:Q&A).
This document covers fundamental concepts in engineering mechanics with a focus on
dynamics, including kinematics, kinetics, particle motion, rigid body motion, and energy
methods. It includes practice problems with solutions to support understanding and exam
preparation. The material is aligned with the 15th edition by Russell C. Hibbeler and is ideal
for engineering students studying dynamics and mechanical systems.
Page 2 of 238
, (Hibbeler 15th dynamics solution manual:Q&A).
© 2022 by R.C. Hibbeler. Published by Pearson Education, Inc., Hoboken, 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:
v = 0
s = 80.7 m
1
Page 3 of 238
, (Hibbeler 15th dynamics solution manual:Q&A).
© 2022 by R.C. Hibbeler. Published by Pearson Education, Inc., Hoboken, 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
Page 4 of 238
, (Hibbeler 15th dynamics solution manual:Q&A).
© 2022 by R.C. Hibbeler. Published by Pearson Education, Inc., Hoboken, 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
2
v = (12 -
v - 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 t == 44 s,s, the
acceleration when the displacement
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
Page 5 of 238
