solution manual for
engineering mechanics: dynamics
by: russell c. hibbeler
15th edition
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.
12–1.
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
dv = (2t - 6) dt
l0 l0
v = t2 - 6t
ds = v dt
s t
ds = (t2 - 6t) dt
l0 l0
t3
s = - 3t 2
3
when t = 6 s,
v = 0 ans.
when t = 11 s,
s = 80.7 m ans.
ans:
s = 80.7 m
2
,12–2.
if a particle has an initial velocity of v 0 = 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 s= s0 + v0 t + a t2
c
2
1
= 0 + 12(10) + ( -2)(10)2
2
= 20 ft ans.
ans:
s = 20 ft
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.
12–3.
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 - 3t2 (1)
dv
a = = -6t t=4 = -24 m>s2 ans.
dt
s t t
ds = v dt = (12 - 3t2)dt
l-10 l1 l1
s + 10 = 12t - t 3 - 11
s = 12t - t3 - 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
δs = - 880 m
st = 912 m
4
engineering mechanics: dynamics
by: russell c. hibbeler
15th edition
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.
12–1.
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
dv = (2t - 6) dt
l0 l0
v = t2 - 6t
ds = v dt
s t
ds = (t2 - 6t) dt
l0 l0
t3
s = - 3t 2
3
when t = 6 s,
v = 0 ans.
when t = 11 s,
s = 80.7 m ans.
ans:
s = 80.7 m
2
,12–2.
if a particle has an initial velocity of v 0 = 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 s= s0 + v0 t + a t2
c
2
1
= 0 + 12(10) + ( -2)(10)2
2
= 20 ft ans.
ans:
s = 20 ft
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.
12–3.
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 - 3t2 (1)
dv
a = = -6t t=4 = -24 m>s2 ans.
dt
s t t
ds = v dt = (12 - 3t2)dt
l-10 l1 l1
s + 10 = 12t - t 3 - 11
s = 12t - t3 - 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
δs = - 880 m
st = 912 m
4
