Engineering Mechanics Dynamics 15th Edition By Russell Hibbeler (Solution Manual)

Engineering
Mechanics, Dynamics,
15e Russell Hibbeler

(Solution Manual all Chapters)


(For Complete File, Download link at the end
of this File)

,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

,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
+2 1 2
1S s = s0 + v0 t + a t
2 c

1
= 0 + 12(10) + ( - 2)(10)2
2

= 20 ft Ans.




Ans:
s = 20 ft

2

, 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
acceleration when
the acceleration when t t == 44 s,s, the
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
ds = v dt = ( 12 - 3t 2 ) dt
L-10 L1 L1

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