PHYS 230
Fall 2007
Assignment 5
Due: 5:00 pm Thursday October 25
1. Two cars are traveling, one behind the other, on a twolane straight road. Each has a
speed of 70 ft/s (about 50 mph) and the distance between them is 90 ft. The driver of
the rear car decides to overtake the car ahead, and does so by accelerating at 6 ft/s
2
up to 100 ft/s (about 70 mph) after which he continues at this speed until he is 90 ft
ahead of the other car. At this time he has also moved back into his own lane.
(a) How far does the overtaking car travel along the road between the beginning and
end of this operation?
Do this part of the problem by working in the moving frame of reference of the
car which is being overtaken. No credit will be given for working in the rest frame
of the road!
First, define some notation. Let car A be the rear car; Car B, the car ahead. So,
initially,
v
A
=
v
B
= 70
ft/s. However, in the rest frame of Car B: these initial
velocities are:
v
A
i
=
v
B
= 0
ft/s.
In this frame, the final velocity of car A is
v
A
f
= 30
ft/s, and the distance that it travels is
d
= 180
ft.
The acceleration
a
= 6
ft/s
2
has the same value in both frames.
In the rest frame of B, the usual kinematic equations relate time, distance and
acceleration. So, the time for A to accelerate, and the distance traveled while it
does so are:
t
1
=
v
A
f

v
A
i
a
= 5 s
,
d
1
=
1
2
at
2
1
= 75 ft
.
The time to travel the remaining distance is:
t
2
=
d

d
1
v
A
f
=
180

75
30
= 3
.
5 s
.
Thus, the total time for the whole manoeuvre is
t
=
t
1
+
t
2
= 8
.
5 s
,
In this time, car B travels along the road through a distance
d
B
=
v
B
t
= 595 ft
and car A travels along the road a distance
d
A
=
d
B
+
d
= 595 + 180 = 775 ft
.
1
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(b) If a third car were in sight, coming in the opposite direction at 88 ft/s (60 mph),
what would be the minimum safe distance between the third car and the overtak
ing car at the beginning of the overtaking operation?
Which frame of reference is most useful for this part of the problem?
It is simplest to do this in the rest frame of the road. (But it could be done either
in the frame of car A or of car C.) In this frame, in the total time
t
, car C travels
a distance
d
C
=
v
C
t
= 748 ft
so that the safe distance wil be
d
safe
=
d
A
+
d
C
= 775 + 748 = 1523 ft
.
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 Fall '07
 Harris
 Energy, 75 ft, 30 ft, 0 Ft, 1 1 k, 70 mph

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