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6. REASONING The distance traveled by the Space Shuttle is equal to its speed multiplied
by the time. The number of football fields is equal to this distance divided by the length L of
one football field.
SOLUTION The number of football fields is ( )( −3 ) 7.6 ×10 m / s 110 ×10 s
x vt
=
= 9.1
Number = =
LL
91.4 m
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7. 3 REASONING In a race against elGuerrouj, Bannister would run a distance given by his
average speed times the time duration of the race (see Equation 2.1). The time duration of
the race would be elGuerrouj’s winning time of 3:43.13 (223.13 s). The difference
between Bannister’s distance and the length of the race is elGuerrouj’s winning margin.
SOLUTION From the table of conversion factors on the page facing the front cover, we
find that one mile corresponds to 1609 m. According to Equation 2.1, Bannister’s average
speed is
Distance
1609 m
Average speed =
=
Elapsed time 239.4 s
Had he run against elGuerrouj at this average speed for the 223.13s duration of the race,
he would have traveled a distance of 1609 m Distance = Average speed × Time = ( 223.13 s ) 239.4 s while elGuerrouj traveled 1609 m. Thus, elGuerrouj would have won by a distance of 1609 m 1609 m − ( 223.13 s ) = 109 m 239.4 s 8. REASONING The younger (and faster) runner should start the race after the older runner,
the delay being the difference between the time required for the older runner to complete the
race and that for the younger runner. The time for each runner to complete the race is equal
to the distance of the race divided by the average speed of that runner (see Equation 2.1).
SOLUTION The difference between the times for the two runners to complete the race is
t50 − t18 , where 48 KINEMATICS IN ONE DIMENSION t50 = Distance
( Average Speed )50yrold and t18 = Distance
( Average Speed )18yrold (2.1) The difference between these two times (which is how much later the younger runner should
start) is
Distance
Distance
t50 − t18 =
−
( Average Speed )50yrold
( Average Speed )18yrold 10.0 × 103 m
10.0 ×103 m
=
−
= 64 s
4.27 m/s
4.39 m/s
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9. REASONING In order for the bear to catch the tourist over the distance d, the bear must
reach the car at the same time as the tourist. During the time t that it takes for the tourist to
reach the car, the bear must travel a total distance of d + 26 m. From Equation 2.1,
vtourist = d
t (1) and vbear = d + 26 m
t (2) Equations (1) and (2) can be solved simultaneously to find d.
SOLUTION Solving Equation (1) for t and substituting into Equation (2), we find
vbear = d + 26 m (d + 26 m)vtourist
=
d / vtourist
d 26 m vbear = 1 +
v
d tourist Solving for d yields: 26 m
26 m
=
= 52 m
vbear
6.0 m/s
−1
−1
4.0 m/s
vtourist
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d= 10. REASONING AND SOLUTION Let west be the positive direction. The average velocity
of the backpacker is x +x
e
v= w
t +t
we where t x
=w
wv
w and x
t=e
ev
e Chapter 2 Problems 49 Combining these equations and solving for xe (suppressing the units) gives
xe = – (1 – v / vw ) xw (1 – v / ve ) = – 1 – (1.34 m/s ) / ( 2.68 m/s ) ( 6.44 km ) 1 – (1.34 m/s ) / ( 0.447 m/s ) = –0.81 km The distance traveled is the magnitude of xe, or 0.81 km .
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11. SSM WWW REASONING Since the woman runs for a known distance at a known
constant speed, we can find the time it takes for her to reach the water from Equation 2.1.
We can then use Equation 2.1 to determine the total distance traveled by the dog in this
time. SOLUTION The time required for the woman to reach the water is
Elapsed time = d woman
vwoman 4.0 km 1000 m = = 1600 s 2.5 m/s 1.0 km In 1600 s, the dog travels a total distance of ddog = vdog t = (4.5 m/s)(1600 s) = 7.2 × 103 m
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12. REASONING
The definition of average velocity is given by Equation 2.2 as
Average velocity = Displacement/(Elapsed time). The displacement in this expression is the
total displacement, which is the sum of the displacements for each part of the trip.
Displacement is a vector quantity, and we must be careful to account for the fact that the
displacement in the first part of the trip is north, while the displacement in the second part is
south.
SOLUTION According to Equation 2.2, the displacement for each part of the trip is the
average velocity for that part times the corresponding elapsed time. Designating north as
the positive direction, we find for the total displacement that
Displacement = m/s
( 27 m/s ) t 3 + 1442444
( −174 ) tSouth
144 North
244
3
Northward Southward where tNorth and tSouth denote, respectively, the times for each part of the trip. Note that the
minus sign indicates a direction due south. Noting that the total elapsed time is
tNorth + tSouth, we can use Equation 2.2 to find the average velocity for the entire trip as
follows: 50 KINEMATICS IN...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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