226
Chapter 2
E
XECUTE
:
(a)
For
0
t
=
to 5.0 s,
0
0
0
30.0 m/s
(5.0 m/s)
75.0 m
2
2
x
x
v
v
x
x
t
+
+
⎛
⎞
⎛
⎞
−
=
=
=
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
. The ball travels a
distance of 75.0 m. For
5.0 s
t
=
to 20.0 s,
0
20.0 m/s
0
(15.0 m/s)
150.0 m
2
x
x
−
+
⎛
⎞
−
=
= −
⎜
⎟
⎝
⎠
. The total distance
traveled is 75.0 m
150.0 m
225.0 m
+
=
.
(b)
The total displacement is
0
75.0 m +(
150.0 m)
75.0 m
x
x
−
=
−
= −
. The ball ends up 75.0 m in the negative
x

direction from where it started.
(c)
For
0
t
=
to 5.0 s,
2
30.0 m/s
0
6.00 m/s
5.0 s
x
a
−
=
=
. For
5.0 s
t
=
to 20.0 s,
2
0
(
20.0 m/s)
1.33 m/s
15.0 s
x
a
− −
=
= +
.
The graph of
x
a
versus
t
is given in Figure 2.64.
(d)
The ball is in contact with the floor for a small but nonzero period of time and the direction of the velocity
doesn’t change instantaneously. So, no, the actual graph of
( )
x
v
t
is not really vertical at 5.00 s.
E
VALUATE
:
For
0
t
=
to 5.0 s, both
x
v
and
x
a
are positive and the speed increases. For
5.0 s
t
=
to 20.0 s,
x
v
is
negative and
x
a
is positive and the speed decreases. Since the direction of motion is not the same throughout, the
displacement is not equal to the distance traveled.
Figure 2.64
2.65.
I
DENTIFY
and
S
ET
U
P
:
Apply constant acceleration equations.
Find the velocity at the start of the second 5.0 s; this is the velocity at the end of the first 5.0 s. Then find
0
x
x
−
for
the first 5.0 s.
E
XECUTE
:
For the first 5.0 s of the motion,
0
0,
x
v
=
5.0 s.
t
=
0
x
x
x
v
v
a t
=
+
gives
(5.0 s).
x
x
v
a
=
This is the initial speed for the second 5.0 s of the motion. For the second 5.0 s:
0
(5.0 s),
x
x
v
a
=
5.0 s,
t
=
0
150 m.
x
x
−
=
2
1
0
0
2
x
x
x
x
v t
a t
−
=
+
gives
2
2
150 m
(25 s
)
(12.5 s
)
x
x
a
a
=
+
and
2
4.0 m/s
x
a
=
Use this
x
a
and consider the first 5.0 s of the motion:
2
2
2
1
1
0
0
2
2
0
(4.0 m/s
)(5.0 s)
50.0 m.
x
x
x
x
v t
a t
−
=
+
=
+
=
E
VALUATE
:
The ball is speeding up so it travels farther in the second 5.0 s interval than in the first. In fact,
0
x
x
−
is proportional to
t
2
since it starts from rest. If it goes 50.0 m in 5.0 s, in twice the time (10.0 s) it should go
four times as far. In 10.0 s we calculated it went 50 m
150 m
200 m,
+
=
which is four times 50 m.
2.66.
I
DENTIFY
:
Apply
2
1
0
0
2
x
x
x
x
v t
a t
−
=
+
to the motion of each train. A collision means the front of the passenger
train is at the same location as the caboose of the freight train at some common time.
S
ET
U
P
:
Let P be the passenger train and F be the freight train. For the front of the passenger train
0
0
x
=
and for
the caboose of the freight train
0
200 m
x
=
. For the freight train
F
15.0 m/s
v
=
and
F
0
a
=
. For the passenger train
P
25.0 m/s
v
=
and
2
P
0.100 m/s
a
= −
.
E
XECUTE
:
(a)
2
1
0
0
2
x
x
x
x
v t
a t
−
=
+
for each object gives
2
1
P
P
P
2
x
v t
a t
=
+
and
F
F
200 m
x
v t
=
+
. Setting
P
F
x
x
=
gives
2
1
P
P
F
2
200 m
v t
a t
v t
+
=
+
.
2
2
(0.0500 m/s
)
(10.0 m/s)
200 m
0
t
t
−
+
=
. The
quadratic formula gives
(
)
2
1
10.0
(10.0)
4(0.0500)(200)
s
(100
77.5) s
0.100
t
=
+
±
−
=
±
. The collision occurs at
100 s
77.5 s
22.5 s
t
=
−
=
. The equations that specify a collision have a physical solution (real, positive
t
), so a
collision does occur.
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 Spring '06
 Buchler
 Physics, Velocity, m/s

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