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Chapter 1: Introduction to Physics
Answers to Even-Numbered Conceptual Questions
2. The quantity T + d does not make sense physically, because it adds together variables that have different
physical dimensions. The quantity d/T does make sense, however; it could represent the distance d traveled
by an object in the time T.
Solutions to Problems
- Picture the Problem : This is a dimensional analysis question.
Strategy: Manipulate the dimensions in the same manner as algebraic expressions.
Solution: Substitute dimensions for the
variables:
2
2
[L]
[L][T]
[T]
[T] [T] therefore 2
p
p
p
a x t
p
Insight: The number 2 does not contribute any dimensions to the problem.
- Picture the Problem : The volume of the oil is spread out into a slick that is one molecule thick.
Strategy: The volume of the slick equals its area times its thickness. Use this fact to find the area.
Solution: Calculate the area for the known volume
and thickness:
3
6 2
6
1.0 m 1 m
2.0 10 m
0.50 m 1 10 m
V
A
h
Insight: Two million square meters is about 772 square miles!
- Picture the Problem : The Earth is roughly a sphere rotating about its axis.
Strategy: Use the fact the Earth spins once about its axis every 24 hours to find the estimated quantities.
Solution: 1. (a) Divide distance by time:
3
3000 mi
1000 mi/h 10 mi/h
3 h
d
v
t
2. (b) Multiply speed by 24 hours:
4
circumference vt 3000 mi/h 24 h 24, 000 mi 10 mi
3. (c) Circumference equals 2 r :
3
circumference 24, 000 mi
3800 mi 10 mi
r
Insight: These estimates are “in the ballpark.” The speed of a point on the equator is 1038 mi/h, the circumference of
the equator is 24,900 mi, and the equatorial radius of the Earth is 3963 mi.
- Picture the Problem : This is a units conversion problem.
Strategy: Multiply the known quantity by appropriate conversion factors to change the units.
Solution: 1. (a) Convert m/s to mi/h:
m 1 mi 3600 s mi
s 1609 m 1 h h
2. (b) Convert m/s to m/ms
3
m 1 10 s m
140 0.14 5.0 ms 0.70 m
s 1 ms ms
Insight: Conversion factors are conceptually equal to one, even though numerically they often equal something other
than one. They are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend fashion.
- Picture the Problem : This is a dimensional analysis question.
Strategy: Find q to make the time dimensions match and then p to make the distance dimensions match. Recall L must
have dimensions of meters and g dimensions of m/s
2
Solution: 1. Make the time dimensions match:
2 1
2 2
L
[T] L = L L [T] implies
[T]
q
q
p p
q
2. Now make the distance units match:
1
2
1
2 2
[L]
[T] L implies
[T]
p
p
Insight: Sometimes you can determine whether you’ve made a mistake in your calculations simply by checking to
ensure the dimensions work out correctly on both sides of your equations.
Chapter 2: One-Dimensional Kinematics
Answers to Even-Numbered Conceptual Questions
8. Yes. For example, your friends might have backed out of a parking place at some point in the trip, giving a
negative velocity for a short time.
16. When she returns to her original position, her speed is the same as it was initially; that is, 4.5 m/s. 18. (a) No. Displacement is the change in position, and therefore it is independent of the location chosen for the
origin. (b) Yes. In order to know whether an object’s displacement is positive or negative, we need to know
which direction has been chosen to be positive.
Solutions to Problems
- Picture the Problem : The dog continuously runs back and forth as
the owners close the distance between each other.
Strategy: First find the time that will elapse before the owners
meet each other. Then determine the distance the dog will cover if
it continues running at constant speed over that time interval.
Solution: 1. Find the time it takes each owner to walk
5.00 m before meeting each other: av
5.00 m
3.8 s
1.3 m/s
d
t
s
2. Find the distance the dog runs:
d s t 3.0 m/s 3.8 s 11 m
Insight: The dog will actually run a shorter distance than this, because it is impossible for it to maintain the same 3.0 m/
s as it turns around to run to the other owner. It must first slow down to zero speed and then accelerate again.
3. (b) For a constant acceleration the velocity varies linearly with time. Therefore we expect the velocity to be equal to
12.5 m/s after half the time (3.11 s) has elapsed.
4. (c) Calculate half the sum of the velocities:
1 1
av,1 2 0 2
v v v 0 12.5 m/s 6.25 m/s
5. Calculate half the sum of the velocities:
1 1
av,2 2 0 2
v v v 12.5 25.0 m/s 18.8 m/s
6. (d) Use the average velocity to find the distance:
1 av,
d v t 6.25 m/s 3.11 s 19.4 m
7. Use the average velocity to find the distance:
2 av,
d v t 18.8 m/s 3.11 s 58.5 m
Insight: The distance traveled is always the average velocity multiplied by the time. This stems from the definition of
average velocity.
- Picture the Problem : The meteorite accelerates from a high speed to rest after impacting the car.
Strategy: Employ the relationship between acceleration, displacement, and velocity (equation 2-12) to find the
acceleration.
Solution: Solve equation 2-12 for acceleration:
2 2 2 2
0 4 2
0 130 m/s
3.8 10 m/s
2 2 0.22 m
v v
a
x
Insight: The high stiffness of steel is responsible for the tremendous (negative) acceleration of the meteorite.
- Picture the Problem : You drive in a straight line and then slow down to a stop.
Strategy: Use the constant acceleration equation of motion (equation 2-7) to find the time. Once the time is known, we
can use the same equation to find the speed. In this case, the acceleration is negative because the car is slowing down.
Solution: 1. (a) Solve equation 2-7 for t :
0
2
0 16 m/s
5.0 s
3.2 m/s
v v
t
a
2. (b) Since the velocity varies linearly with time for constant acceleration, the velocity will be half the initial velocity
when you have braked for half the time. Therefore the speed after braking 2.5 s will be equal to 8.0 m/s.
3. Evaluate equation 2-7 directly:
2
0
v v at 16 m/s 3.2 m/s 2.5 s 8.0 m/s
4. (c) The total distance traveled is the distance the
car travels at 16 m/s before you hit the brakes (a time
interval given by your reaction time) plus the distance
covered as the car stops.
0 react av stop
av stop
react
0
55 m 8.0 m/s 5.0 s
0.94 s
16 m/s
x v t v t
x v t
t
v
Insight: For constant acceleration, the velocity changes linearly with time, but nonlinearly with distance.
- Picture the Problem : The apple falls straight downward under the influence of gravity.
Strategy: The distance of the fall is estimated to be about 3.0 m (about 10 ft). Then use the time-free equation of motion
(equation 2-12) to estimate the speed of the apple.
Solution: 1. Solve equation 2-12 for v , assuming
the apple drops from rest ( 0
v 0
):
v 0 2 a x
2. Let a = g and calculate v :
2
v 2 9.81 m/s 3.0 m 7.7 m/s 17 mi/h
Insight: Newton supposedly then reasoned that the same force that made the apple fall also keeps the Moon in orbit
around the Earth, leading to his universal law of gravity (Chapter 12). One lesson we might learn here is—wear a helmet
when sitting under an apple tree!
