Preparation problems
2-1. In 1994 Leroy Burrell of the United States set a
world record in the 100 m dash, with a time of 9.84 s. What was his
average speed?
Solution:
From our constants and equation sheet, we see
or in this case, since we are dealing with average quantities,
Taking Dx = 100 m and Dt = 9.84 s, we see
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vav = |
100 m
9.84 s
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= 10.16 m/s. |
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2-5. Starting from home, you bicycle 24 km north in
2.5 h, then turn around and pedal straight home in 1.5 h. What are your
(a) displacement at the end of the first 2.5 h, (b)
average velocity over the first 2.5 h, (c) displacement for the
entire trip, and (d) average velocity for the entire trip?
Solution:
(a) At the end of the first 2.5 h, it is clear you have travelled
Dx = 24 km = 24,000 m.
(b) Recalling from the previous problem that
we see here that Dx = 24,000 m and
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Dt = 2.5 h · |
3600 s
1 h
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= 9000 s. |
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Using these numbers, we see
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vav = |
24,000 m
9000 s
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= 2.67 m/s. |
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(c) At the end of the trip, you are back home,
therefore, your total displacement must be 0 m.
(d) Since Dx = 0 m, equation (1) immediately
tells us that vav = 0 m/s.
2-18. Figure 2-18 (on p. 41) shows the position of an
object as a function of time. Determine the average velocity for
(a) the first 2 s; (b) the first 4 s; (c) the first
6 s; and (d) the interval from 3 s to 4 s.
Solution:
(a) We begin again with the equation
To solve each portion of the problem, we need to find the starting and
ending points of the time interval given. For the first 2 s, we have
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vav(0 ® 2 s) = |
x(2 s) - x0
2 s - 0 s
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= |
2.5 m - 0 m
2 s
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= 1.25 m/s. |
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(b) Again we start with equation (1), only with x(4
s) = 0 m. This gives us
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vav(0 ® 4 s) = |
x(4 s) - x0
4 s - 0 s
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= |
0 m - 0 m
4 s
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= 0 m/s. |
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(c) Again we start with equation (1), only with x(6
s) = -2 m. This gives us
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vav(0 ® 6 s) = |
x(6 s) - x0
6 s - 0 s
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= |
-2 m - 0 m
6 s
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= -0.33 m/s. |
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(c) Finally, we start with equation (1), only with
x(3 s) = 3 m and x(4 s) = 0 m . This gives us
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vav(3 ® 4 s) = |
x(4 s) - x(3 s)
4 s - 3 s
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= |
0 m - (-3 m)
1 s
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= -3 m/s. |
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Follow-up problems
1-25. An equation you'll encounter in the next chapter
is x = 1/2 a t2, where x is distance and t is time. Use
dimensional analysis to find the dimensions of a.
Solution:
We start by examining the dimensions of what we know on each
side of the equation
We know the constant 1/2 is dimensionless,
while the other items in the equation have the dimnsions:
So the left side of the equation (1) has dimensions of length. This means
the right side must also have the dimensions of length. So
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[length] = [time]2 [dimensions of a] . |
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Dividing both sides by [time]2, we get our result:
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[dimensions of a] = |
[length]
[time]2
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. |
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1-47. (a) Estimate the volume of water in
Earth's oceans. (b) If scientists succeed in harnessing nuclear
fusion as an energy source, each gallon of sea water will be equivalent
to 340 gallons of gasoline. At our present rate of gasoline consumption
(given as between 5 ×1010 and 1 ×1011 gallons/year
in Example 1-6), how long would the oceans supply our fuel needs?
Solution:
(a) There are many ways to approach an estimation
problem, but we will focus on only one. We will find the volume of the
oceans by assuming that it is given by the simple formula
where V = volume of the oceans, A = the surface area of the oceans,
and [`d] = the average depth of the ocean.
We can find the surface area of the ocean by noting that between 5/8 and
2/3 of Earth's surface is covered by water-we will use 5/8 = 0.6. So
recalling how to calculate the surface area of a sphere, we see
where Re = 6.37 ×106 m.
Now we need to determine the average depth of the oceans. We know this
must be somewhere between the deepest point (the Marianas Trench in the
Pacific, which is about 11,000 m deep) and sea level. A reasonable
guess might be 1000 m (the actual number is around 4000 m), so that is
what we will use.
Putting this all together, we get one possible estimate of
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(0.6) (4 p) (6.37 ×106 m)2 ·(1000 m) |
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(b) We must now use our answer from part (a) to
determine the energy stored in the oceans, and how long this should
supply our energy needs. We can proceed several ways. First recall
the conversion
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1 gallon = 3.785 ×10-3 m3 . |
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Next note that the fuel in one gallon of water is equal to 340 gallons
of gasoline, and finally, recall that the rate of gasoline consumption
is 1 ×1011 gallons of gasoline/year.
Putting this information all together, we can find the time to use all
the water in the oceans by finding the equivalent gasoline content of
the oceans and dividing by the consumption rate. In equation form, this
looks like
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1 [m3 water] » 9 ×104 [gallons gasoline] . |
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Using this conversion, we quickly see that the total energy content of
the oceans will last for
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( m3 water ) · |
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m3 gas
m3 water
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gal gas
m3 gas
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( 3 ×1017 m3 water ) · |
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340 |
m3 gas
m3 water
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1 gal gas
3.785 ×10-3 m3 gas
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Solutions translated from TEX by TTH, version 1.57.
