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Engineering Mechanics  Dynamics
Chapter 12
Problem 121
A truck traveling along a straight road at speed
v
1
, increases its speed to
v
2
in time
t
. If its
acceleration is constant, determine the distance traveled.
Given:
v
1
20
km
hr
v
2
120
km
hr
t
15 s
Solution:
a
v
2
v
1
0
t
a
1.852
m
s
2
d
v
1
t
1
2
at
2
.
d
291.67 m
Problem 122
A car starts from rest and reaches a speed
v
after traveling a distance
d
along a straight road.
Determine its constant acceleration and the time of travel.
Given:
v
80
ft
s
d
500 ft
Solution:
v
2
2
ad
a
v
2
2
d
a
6.4
ft
s
2
v
t
v
a
t
12.5 s
Problem 123
A baseball is thrown downward from a tower of height
h
with an initial speed
v
0
. Determine
the speed at which it hits the ground and the time of travel.
Given:
h
50 ft
g
32.2
ft
s
2
v
0
18
ft
s
Solution:
v
v
0
2
2
gh
.
v
59.5
ft
s
1
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View Full DocumentEngineering Mechanics  Dynamics
Chapter 12
t
v
v
0
0
g
t
1.29 s
*
Problem 12–4
Starting from rest, a particle moving in a straight line has an acceleration of
a
= (
bt
+
c
). What
is the particle’s velocity at
t
1
and what is its position at
t
2
?
Given:
b
2
m
s
3
c
6
0
m
s
2
t
1
6 s
t
2
11 s
Solution:
a t
( )
bt
c
.
v t
( )
0
t
t
a t
( )
´
µ
¶
d
d t
( )
0
t
t
v t
( )
´
µ
¶
d
v t
1
+ ,
0
m
s
d t
2
+ ,
80.7 m
Problem 125
Traveling with an initial speed
v
0
a car accelerates at rate
a
along a straight road. How long will it
take to reach a speed
v
f
? Also, through what distance does the car travel during this time?
Given:
v
0
70
km
hr
a
6000
km
hr
2
v
f
120
km
hr
Solution:
v
f
v
0
at
.
t
v
f
v
0
0
a
t
30 s
v
f
2
v
0
2
2
as
.
s
v
f
2
v
0
2
0
2
a
s
792 m
Problem 126
A freight train travels at
v
v
0
1
e
b
0
t
0
+ ,
where
t
is the elapsed time. Determine the distance
traveled in time
t
1
, and the acceleration at this time.
2
Engineering Mechanics  Dynamics
Chapter 12
v
0
60
ft
s
b
1
s
t
1
3 s
Solution:
v t
( )
v
0
1
e
b
0
t
0
+ ,
a t
( )
t
v t
( )
d
d
d
t
( )
0
t
t
v t
( )
´
µ
¶
d
d
t
1
+ ,
123.0 ft
a t
1
+ ,
2.99
ft
s
2
Problem 127
The position of a particle along a straight line is given by
s
p
=
at
3
+
bt
2
+
ct
. Determine its
maximum acceleration and maximum velocity during the time interval
t
0
d
t
d
t
f
.
Given:
a
1
ft
s
3
b
9
0
ft
s
2
c
15
ft
s
t
0
0 s
t
f
10 s
Solution:
s
p
at
3
bt
2
.
ct
.
v
p
t
s
p
d
d
3
2
2
.
c
.
a
p
t
v
p
d
d
2
t
s
p
d
d
2
6
2
b
.
Since the acceleration is linear in time then the maximum will occur at the start or at the end.
We check both possibilities.
a
max
max 6
0
b
.
6
f
2
b
.
/
+ ,
a
max
42
ft
s
2
The maximum velocity can occur at the beginning, at the end, or where the acceleration is zero.
We will check all three locations.
t
cr
b
0
3
a
t
cr
3 s
3
Given:
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View Full DocumentEngineering Mechanics  Dynamics
Chapter 12
v
max
max 3
at
0
2
2
bt
0
.
c
.
3
f
2
2
f
.
c
.
/
3
cr
2
2
cr
.
c
.
/
+ ,
v
max
135
ft
s
*
Problem 128
From approximately what floor of a building must a car be dropped from an atrest position
so that it reaches a speed
v
f
when it hits the ground? Each floor is a distance
h
higher than the
one below it. (Note: You may want to remember this when traveling at speed
v
f
)
Given:
v
f
55 mph
h
12 ft
g
32.2
ft
s
2
Solution:
a
c
g
v
f
2
0
2
a
c
s
.
H
v
f
2
2
a
c
H
101.124 ft
Number of floors
N
Height of one floor
h
12ft
N
H
h
N
8.427
N
ceil
N
( )
The car must be dropped from floor number
N
9
Problem 12–9
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 Spring '08
 Vandenbrink

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