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F10PHYS115UAE
Assignment 1 Solutions
1/4
30.
INTERPRET
The problem states that the acceleration of the car is
constant
, so we can use the constant
acceleration equations and techniques developed in this chapter. We’re given initial and final speeds, and the time,
and we’re asked to find the distance.
DEVELOP
Equation 2.9 relates distance to initial and final speed and to time—that’s just what we need. We need
the distance traveled during the given time, which would be the difference between
x
and
.
o
x
We also need to be
careful of our units—the problem gives us speeds in km/h and time in seconds, so we have to convert so that
everything is in the same units.
EVALUATE
First, convert the speeds to units of m/s.
70 km
1000 m
1 hour
1 hour
1 km
3600 seconds
19.4 m/s.
Similarly, 80 km/h
22.2 m/s.
11
22
()
oo
o
o
x
xv
v
t
x
x
v
v
t
1m
m
(
)
19.4
22.2
(6 s)
125 m
2s
s
o
xx
ASSESS
These speeds and the time are consistent with passing on a country road. The distance is about right for a
passing distance under these circumstances, too; so the answer seems reasonable.
45.
INTERPRET
This is a onedimensional problem involving two travel segments.
DEVELOP
The trip can be divided into two time intervals,
1
t
and
2
t
with
12
40 min (2 h/3).
ttt
The total
distance traveled is
x
x
x
25 mi, where
1
x
and
2
x
are the distances covered in each time interval.
EVALUATE
During the first time interval,
1
t
15 min (or 0.25 h), with an average speed of
1
v
20 mi/h, the
distance traveled is
1
(20 mi/h)(0.25 h)
5 mi
t
Therefore, the remaining distance
21
25 mi 5mi
20 mi
x
must be covered in
5
40 min 15 min
25 min
h
12
tt
t
This implies an average speed of
2
2
20 mi
2
5 h/12
48 mi/h.
x
t
v
ASSESS
The overall average speed was predetermined to be
25 mi
37.5 mi/h
2 h/3
x
v
t
When you drive slower during the first segment, you make it up by driving faster during the second. In fact, the
overall average speed equals the timeweighted average of the average speeds for the two parts of the trip:
12 1
2
1
2
15 min
25 min
(20 mi/h)
(48 mi/h)
37.5 mi/h
40 min
40 min
xx v
tv
t
t
t
x
vv
v
t
t
t
58.
(
a
) and (
b
) The train goes from velocity
0
110 km/h
30.6 m/s
v
(positive eastward) at
0
0,
t
to a stop,
0,
v
at
1.2 min
72 s.
t
The constant acceleration was
2
00
(
)/(
)
(30.6 m/s)/(72 s)
0.424
m/s .
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This note was uploaded on 10/05/2010 for the course PHYS 115 taught by Professor Rohan during the Fall '06 term at Waterloo.
 Fall '06
 ROHAN
 mechanics, Acceleration

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