DEPARTMENT OF MECHANICAL ENGINEERING
Faculty of Engineering & Applied Sciences
OFFICIAL COURSE MATERIAL · FOR ENROLLED STUDENTS ONLY
EME
DYNAMICS
OFFICIAL SOLUTIONS MANUAL
Engineering Mechanics
DYNAMICS
Twelfth Edition
R. C. Hibbeler
COURSE INFORMATION
Course: ME 302 — Engineering Dynamics
Semester: Spring 2026
Instructor: Professor J. A. Smith, Ph.D., P.E.
Pages: 1 – 250 (Part I of II)
CONFIDENTIAL — NOT FOR DISTRIBUTION — INSTRUCTOR USE ONLY
© 2012 Pearson Education, Inc. · Upper Saddle River, NJ · All Rights Reserved
, 91962_01_s12-p0001-0176 6/8/09 8:05 AM Page 1
ENGINEERING MECHANICS: DYNAMICS — 12th Ed. | R. C. Hibbeler SOLUTIONS MANUAL | Spring 2026
© 2010 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. A car starts from rest and with constant
acceleration achieves a velocity of 15 m>s when it travels a
distance of 200 m. Determine the acceleration of the car
and the time required.
Kinematics:
v0 = 0, v = 15 m>s, s0 = 0, and s = 200 m.
A:
+ B v2 = v0 2 + 2ac(s - s0)
152 = 02 + 2ac(200 - 0)
ac = 0.5625 m>s2 Ans.
A:
+ B v = v0 + act
15 = 0 + 0.5625t
t = 26.7 s Ans.
12–2. A train starts from rest at a station and travels with
a constant acceleration of 1 m>s2. Determine the velocity of
the train when t = 30 s and the distance traveled during
this time.
Kinematics:
ac = 1 m>s2, v0 = 0, s0 = 0, and t = 30 s.
A:
+ B v = v0 + act
= 0 + 1(30) = 30 m>s Ans.
A:
+ B 1 2
s = s0 + v0t + at
2 c
(1) A 302 B
1
= 0 + 0 +
2
= 450 m Ans.
1
Department of Mechanical Engineering · Confidential Page 1 of 250
, 91962_01_s12-p0001-0176 6/8/09 8:05 AM Page 2
ENGINEERING MECHANICS: DYNAMICS — 12th Ed. | R. C. Hibbeler SOLUTIONS MANUAL | Spring 2026
© 2010 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. An elevator descends from rest with an acceleration
of 5 ft>s2 until it achieves a velocity of 15 ft>s. Determine the
time required and the distance traveled.
Kinematics:
ac = 5 ft>s2, v0 = 0, v = 15 ft>s, and s0 = 0.
A+TB v = v0 + act
15 = 0 + 5t
t = 3s Ans.
A+TB v2 = v0 2 + 2ac(s - s0)
152 = 02 + 2(5)(s - 0)
s = 22.5 ft Ans.
*12–4. A car is traveling at 15 m>s, when the traffic light
50 m ahead turns yellow. Determine the required constant
deceleration of the car and the time needed to stop the car
at the light.
Kinematics:
v0 = 0, s0 = 0, s = 50 m and v0 = 15 m>s.
A:
+ B v2 = v0 2 + 2ac(s - s0)
0 = 152 + 2ac(50 - 0)
ac = -2.25 m>s2 = 2.25 m>s2 ; Ans.
A:
+ B v = v0 + act
0 = 15 + (-2.25)t
t = 6.67 s Ans.
2
Department of Mechanical Engineering · Confidential Page 2 of 250
, 91962_01_s12-p0001-0176 6/8/09 8:05 AM Page 3
ENGINEERING MECHANICS: DYNAMICS — 12th Ed. | R. C. Hibbeler SOLUTIONS MANUAL | Spring 2026
© 2010 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. A particle is moving along a straight line with the
acceleration a = (12t – 3t1/2) ft>s2, where t is in seconds.
Determine the velocity and the position of the particle as a
function of time. When t = 0, v = 0 and s = 15 ft.
Velocity:
A:
+ B dv = a dt
A 12t - 3t1>2 B dt
v t
dv =
L0 L0
v冷0 = A 6t2 - 2t3>2 B 2
v t
0
v = A 6t2 - 2t3>2 B ft>s Ans.
Position: Using this result and the initial condition s = 15 ft at t = 0 s,
A:
+ B ds = v dt
A 6t2 - 2t3>2 B dt
s t
ds =
L15 ft L0
4 5>2 2 t
s冷15 ft = a 2t3 - t b
s
5 0
s = a 2t3 - t + 15 bft
4 5>2
Ans.
5
12–6. A ball is released from the bottom of an elevator
which is traveling upward with a velocity of 6 ft>s. If the ball
strikes the bottom of the elevator shaft in 3 s, determine the
height of the elevator from the bottom of the shaft at the
instant the ball is released. Also, find the velocity of the ball
when it strikes the bottom of the shaft.
Kinematics: When the ball is released, its velocity will be the same as the elevator at
the instant of release. Thus, v0 = 6 ft>s. Also, t = 3 s, s0 = 0, s = -h, and
ac = -32.2 ft>s2.
A+cB
1
s = s0 + v0t + a t2
2 c
(-32.2) A 32 B
1
-h = 0 + 6(3) +
2
h = 127 ft Ans.
A+cB v = v0 + act
v = 6 + (-32.2)(3)
= -90.6 ft>s = 90.6 ft>s T Ans.
3
Department of Mechanical Engineering · Confidential Page 3 of 250
Faculty of Engineering & Applied Sciences
OFFICIAL COURSE MATERIAL · FOR ENROLLED STUDENTS ONLY
EME
DYNAMICS
OFFICIAL SOLUTIONS MANUAL
Engineering Mechanics
DYNAMICS
Twelfth Edition
R. C. Hibbeler
COURSE INFORMATION
Course: ME 302 — Engineering Dynamics
Semester: Spring 2026
Instructor: Professor J. A. Smith, Ph.D., P.E.
Pages: 1 – 250 (Part I of II)
CONFIDENTIAL — NOT FOR DISTRIBUTION — INSTRUCTOR USE ONLY
© 2012 Pearson Education, Inc. · Upper Saddle River, NJ · All Rights Reserved
, 91962_01_s12-p0001-0176 6/8/09 8:05 AM Page 1
ENGINEERING MECHANICS: DYNAMICS — 12th Ed. | R. C. Hibbeler SOLUTIONS MANUAL | Spring 2026
© 2010 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. A car starts from rest and with constant
acceleration achieves a velocity of 15 m>s when it travels a
distance of 200 m. Determine the acceleration of the car
and the time required.
Kinematics:
v0 = 0, v = 15 m>s, s0 = 0, and s = 200 m.
A:
+ B v2 = v0 2 + 2ac(s - s0)
152 = 02 + 2ac(200 - 0)
ac = 0.5625 m>s2 Ans.
A:
+ B v = v0 + act
15 = 0 + 0.5625t
t = 26.7 s Ans.
12–2. A train starts from rest at a station and travels with
a constant acceleration of 1 m>s2. Determine the velocity of
the train when t = 30 s and the distance traveled during
this time.
Kinematics:
ac = 1 m>s2, v0 = 0, s0 = 0, and t = 30 s.
A:
+ B v = v0 + act
= 0 + 1(30) = 30 m>s Ans.
A:
+ B 1 2
s = s0 + v0t + at
2 c
(1) A 302 B
1
= 0 + 0 +
2
= 450 m Ans.
1
Department of Mechanical Engineering · Confidential Page 1 of 250
, 91962_01_s12-p0001-0176 6/8/09 8:05 AM Page 2
ENGINEERING MECHANICS: DYNAMICS — 12th Ed. | R. C. Hibbeler SOLUTIONS MANUAL | Spring 2026
© 2010 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. An elevator descends from rest with an acceleration
of 5 ft>s2 until it achieves a velocity of 15 ft>s. Determine the
time required and the distance traveled.
Kinematics:
ac = 5 ft>s2, v0 = 0, v = 15 ft>s, and s0 = 0.
A+TB v = v0 + act
15 = 0 + 5t
t = 3s Ans.
A+TB v2 = v0 2 + 2ac(s - s0)
152 = 02 + 2(5)(s - 0)
s = 22.5 ft Ans.
*12–4. A car is traveling at 15 m>s, when the traffic light
50 m ahead turns yellow. Determine the required constant
deceleration of the car and the time needed to stop the car
at the light.
Kinematics:
v0 = 0, s0 = 0, s = 50 m and v0 = 15 m>s.
A:
+ B v2 = v0 2 + 2ac(s - s0)
0 = 152 + 2ac(50 - 0)
ac = -2.25 m>s2 = 2.25 m>s2 ; Ans.
A:
+ B v = v0 + act
0 = 15 + (-2.25)t
t = 6.67 s Ans.
2
Department of Mechanical Engineering · Confidential Page 2 of 250
, 91962_01_s12-p0001-0176 6/8/09 8:05 AM Page 3
ENGINEERING MECHANICS: DYNAMICS — 12th Ed. | R. C. Hibbeler SOLUTIONS MANUAL | Spring 2026
© 2010 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. A particle is moving along a straight line with the
acceleration a = (12t – 3t1/2) ft>s2, where t is in seconds.
Determine the velocity and the position of the particle as a
function of time. When t = 0, v = 0 and s = 15 ft.
Velocity:
A:
+ B dv = a dt
A 12t - 3t1>2 B dt
v t
dv =
L0 L0
v冷0 = A 6t2 - 2t3>2 B 2
v t
0
v = A 6t2 - 2t3>2 B ft>s Ans.
Position: Using this result and the initial condition s = 15 ft at t = 0 s,
A:
+ B ds = v dt
A 6t2 - 2t3>2 B dt
s t
ds =
L15 ft L0
4 5>2 2 t
s冷15 ft = a 2t3 - t b
s
5 0
s = a 2t3 - t + 15 bft
4 5>2
Ans.
5
12–6. A ball is released from the bottom of an elevator
which is traveling upward with a velocity of 6 ft>s. If the ball
strikes the bottom of the elevator shaft in 3 s, determine the
height of the elevator from the bottom of the shaft at the
instant the ball is released. Also, find the velocity of the ball
when it strikes the bottom of the shaft.
Kinematics: When the ball is released, its velocity will be the same as the elevator at
the instant of release. Thus, v0 = 6 ft>s. Also, t = 3 s, s0 = 0, s = -h, and
ac = -32.2 ft>s2.
A+cB
1
s = s0 + v0t + a t2
2 c
(-32.2) A 32 B
1
-h = 0 + 6(3) +
2
h = 127 ft Ans.
A+cB v = v0 + act
v = 6 + (-32.2)(3)
= -90.6 ft>s = 90.6 ft>s T Ans.
3
Department of Mechanical Engineering · Confidential Page 3 of 250
