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MOTION IN A STRAIGHT LINE
E
XERCISES
Section 2.1 Average Motion
13.
I
NTERPRET
We need to find average speed, given distance and time.
D
EVELOP
Speed is distance divided by time.
E
VALUATE
100 m
9.77 s
10.2 m/s
v
=
=
A
SSESS
His time is about 10 seconds, so his speed is about 10 m/s.
14.
I
NTERPRET
We need to find average speed, given distance and time.
D
EVELOP
The Olympic marathon distance is 42,186 m. We convert her time to seconds, and use
x
t
v
=
to find her
speed.
E
VALUATE
3600 s
60 s
1 h
1 min
2 h
26 min
20 s
(2 h)
(26 min)
(20 s)
8780 s.
+
+
=
×
+
×
+
=
42,186 m
4.805 m/s
8780 s
x
v
t
=
=
=
A
SSESS
To check our answer we can convert this speed to mph, and see if it’s reasonable for a distance of about
26 miles in about 2.5 hours.
15.
I
NTERPRET
This is a onedimensional kinematics problem, and we identify the bicyclist as the object of interest.
His trip consists of two parts (out and back), and we are asked to compute the displacement and average velocity of
each part, as well as for the trip as a whole.
D
EVELOP
For motion in a straight line, the displacement, or the net change in position, is
2
1
,
x
x
x
∆
=

where
x
1
and
x
2
are the starting and the end points, respectively. The average velocity is the displacement divided by the
time interval,
/
,
v
x t
= ∆ ∆
as shown in Equation 2.1. In our coordinate system, we take north to be +
x
.
E
VALUATE
(a)
The displacement at the end of the first 2.5 h is
out
2
1
24 km
0 km
24 km
x
x
x
∆
=

=

=
(b)
With
out
2.5 h,
t
∆
=
the average velocity over this interval is
out
out
out
24 km
9.6 km h
(north)
2.5 h
x
v
t
∆
=
=
=
/
∆
(c)
With
back
1.5 h,
t
∆
=
the average velocity for the homeward leg of the trip is
back
back
back
24 km
16 km/h
(south)
1.5 h
wx
v
t
∆

=
=
= 
∆
(d)
Since the final position of the bicyclist is the same as his initial position, his total displacement of the trip is
total
out
back
24 km
( 24 km)
0 km.
x
x
x
∆
= ∆
+ ∆
=
+ 
=
(e)
Since
total
0,
x
∆
=
the average velocity for the entire trip is
total
total
total
/
0.
v
x
t
= ∆
∆
=
A
SSESS
Note the distinction between average velocity and average speed. The former depends only on the net
displacement, while the latter takes into consideration the total distance traveled. The average speed for this trip is
out
back
out
back




24 km
24 km
average speed
12 km/h
2.5 h
1.5 h
x
x
t
t
∆
+ ∆
+
=
=
=
∆
+ ∆
+
2.1
2
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Chapter 2
16.
I
NTERPRET
We must find the time it takes for a radio signal to travel a given distance.
D
EVELOP
Radio waves travel at a constant speed of
8
3.0 1
0
v
=
×
m/s. We
know that the distance is
9
12
1
.2 1
0 km
1.2 1
0 m,
x
=
×
=
×
so we can use the definition of speed,
,
x
t
v
=
to calculate the time.
E
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This note was uploaded on 09/12/2009 for the course PH PH101 taught by Professor Prof.kim during the Spring '09 term at Korea Advanced Institute of Science and Technology.
 Spring '09
 Prof.Kim
 Physics

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