Vật lý đại cương Principles of Physic Chương I

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Principles of Physics: A Calculus-Based Text,
Fifth Edition, International Edition
Raymond A. Serway
John W. Jewett, Jr.
Publisher, Physical Sciences: Mary Finch
Publisher, Astronomy and Physics: Charles Hartford
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4
Introduction and Vectors
1
Chapter
Chapter Outline
1.1 Standards of Length, Mass,
andTime
1.2 Dimensional Analysis
1.3 Conversion of Units
1.4 Order-of-Magnitude Calculations
1.5 Signifi cant Figures
1.6 Coordinate Systems
1.7 Vectors and Scalars

1.8 Some Properties of Vectors
1.9 Components of a Vector and Unit
Vectors
1.10 Modeling, Alternative
Representations, and
Problem-Solving Strategy
SUMMARY
T
he goal of physics is to provide a
quantitative understanding of cer-
tain basic phenomena that occur in our
Universe. Physics is a science based on
experimental observations and mathematical analyses. The main objectives
behind suchexperiments and analyses are to develop theories that explain the
phenomenon being studied and to relate those theories to other established
theories. Fortunately, it is possible to explain the behavior of various physical
systems using relatively few fundamental laws. Analytical procedures require
the expression of those laws in the language of mathematics, the tool that provides a
bridge between theory and experiment. In this chapter, we shall discuss a few math-
ematical concepts and techniques that will be used throughout the text. In addition,
we will outline an effective problem-solving strategy that should be adopted and
used in your problem-solving activities throughout the text.
1.1
|
Standards of Length, Mass, and Time
To describe natural phenomena, we must make measurements associated with
physical quantities, such as the length of an object. The laws of physics can be
expressed as mathematical relationships among physical quantities that will be
Raymond A. Serway
A signpost in Saint Petersburg, Florida,

shows the distance and direction to several
cities. Quantities that are defi ned by both a
magnitude and a direction are called vector
quantities.
Interactive
content from this and other chapters may
be assigned online in Enhanced WebAssign.
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1.1
|
Standards of Length, Mass, and Time5
introduced and discussed throughout the book. In mechanics, the three fundamen-
tal quantities are length, mass, and time. All other quantities in mechanics can be
expressed in terms of these three.
If we measure a certain quantity and wish to describe it to someone, a unit for the
quantity must be speci ed and de ned. For example, it would be meaningless for
a visitor from another planet to talk to us about a length of 8.0 “glitches” if we did
not know the meaning of the unit glitch. On the other hand, if someone familiar
with our system of measurement reports that a wall is 2.0 meters high and our unit
of length is de ned to be 1.0 meter, we then know that the height of the wall is twice
our fundamental unit of length. An international committee has agreed on a system
of de nitions and standards to describe fundamental physical quantities. It is called
the SI system (Système International) of units. Its units of length, mass, and time are
the meter, kilogram, and second, respectively.
Length
In .. 1120, King Henry I of England decreed that the standard of length in his

country would be the yard and that the yard would be precisely equal to the distance
from the tip of his nose to the end of his outstretched arm. Similarly, the original
standard for the foot adopted by the French was the length of the royal foot of King
Louis XIV. This standard prevailed until 1799, when the legal standard of length
in France became the meter, de ned as one ten-millionth of the distance from the
equator to the North Pole.
Many other systems have been developed in addition to those just discussed, but
the advantages of the French system have caused it to prevail in most countries and
in scienti c circles everywhere. Until 1960, the length of the meter was de ned as the
distance between two lines on a speci c bar of platinum–iridium alloy stored under
controlled conditions. This standard was abandoned for several reasons, a principal
one being that the limited accuracy with which the separation be tweenthe lines can
be determined does not meet the current requirements of science and technology.
The de nition of the meter was modi ed to be equal to 1 650 763.73 wavelengths of
orange–red light emitted from a krypton-86 lamp. In October 1983, the meter was
rede ned to be the distance traveled by light in a vacuum during a time interval of
1y299 792 458 second. This value arises from the establishment of the speed of light in
a vacuum as exactly 299 792 458 meters per second. We will use the standard scienti c
notation for numbers with more than three digits in which groups of three digits are
separated by spaces rather than commas. Therefore, 1 650 763.73 and 299 792 458
in this paragraph are the same as the more popular American cultural notations of
1,650,763.73 and 299,792,458. Similarly, ␲ 5 3.14159265 is written as 3.141 592 65.
Mass
Mass represents a measure of the resistance of an object to changes in its motion. The
SI unit of mass, the kilogram, is de ned as the mass of a speci c platinum–iridium
alloy cylinder kept at the International Bureau of Weights and Measures at Sèvres,
France. At this point, we should add a word of caution. Many beginning students of
physics tend to confuse the physical quantities called weight and mass. For the pres-
ent we shall not discuss the distinction between them; they will be clearly de ned in
later chapters. For now you should note that they are distinctly different quantities.

Time
Before 1967, the standard of time was de ned in terms of the average length of a mean
solar day. (A solar day is the time interval between successive appearances of the Sun
at the highest point it reaches in the sky each day.) The basic unit of time, the sec-
ond, was de ned to be (1/60)(1/60)(1/24) 5 1/86 400 of a mean solar day. In 1967,
the second was rede ned to take advantage of the great precision obtainable with a
device known as an atomic clock (Fig. 1.1), which uses the characteristic frequency of the
c
De nition of the meter
c
De nition of the kilogram
Figure 1.1 A cesium fountain
atomic clock. The clock will neither
gain nor lose a second in 20 million
years.
© 2005 Geoffrey Wheeler Photography
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6CHAPTER 1
|
Introduction and Vectors
cesium-133 atom as the “reference clock.” The second is now de ned as 9 192 631 770
times the period of oscillation of radiation from the cesium atom. It is possible today
to purchase clocks and watches that receive radio signals from an atomic clock in
Colorado, which the clock or watch uses to continuously reset itself to the correct time.
Approximate Values for Length, Mass, and Time
Approximate values of various lengths, masses, and time intervals are presented in

Tables 1.1, 1.2, and 1.3, respectively. Note the wide range of values for these quantities.
1

You should study the tables and begin to generate an intuition for what is meant by a
mass of 100 kilograms, for example, or by a time interval of 3.2 3 10
7
seconds.
Systems of units commonly used in science, commerce, manufacturing, and
everyday life are (1) the SI system, in which the units of length, mass, and time are the
meter (m), kilogram (kg), and second (s), respectively; and (2) the U.S. customary
system, in which the units of length, mass, and time are the foot (ft), slug, and second,
respectively. Throughout most of this text we shall use SI units because they are
almost universally accepted in science and industry. We will make limited use of U.S.
customary units in the study of classical mechanics.
Some of the most frequently used pre xes for the powers of ten and their abbreviations
are listed in Table 1.4. For example, 10
23
m is equivalent to 1 millimeter (mm), and 10
3
m
is 1 kilometer (km). Likewise, 1 kg is 10
3
grams (g), and 1 megavolt (MV) is 10
6
volts (V).
The variables length, time, and mass are examples of fundamental quantities. A much
larger list of variables contains derived quantities, or quantities that can be expressed as
a mathematical combination of fundamental quantities. Common examples are area,
which is a product of two lengths, and speed, which is a ratio of a length to a time interval.
Another example of a derived quantity is density. The density ␳ (Greek letter rho;

a table of the letters in the Greek alphabet is provided at the back of the book) of any
substance is de ned as its mass per unit volume:

␳
;
m
V
1.1
b
c
De nition of density
Pitfall Prevention | 1.1
Reasonable Values
Generating intuition about typical
values of quantities when solving
problems is important because you
must think about your end result and
determine if it seems reasonable. For
example, if you are calculating
the mass of a house y and arrive
at a value of 100 kg, this answer is
unreasonable and there is an error
somewhere.
1
If you are unfamiliar with the use of powers of ten (scienti c notation), you should review Appendix B.1.
TABLE 1.1
|
Approximate Values of Some Measured Lengths
Length (m)
Distance from the Earth to the most remote quasar 1.4 3 10

26
Distance from the Earth to the most remote normal galaxies 9 3 10
25
Distance from the Earth to the nearest large galaxy (M 31, the Andromeda galaxy) 2 3 10
22
Distance from the Sun to the nearest star (Proxima Centauri) 4 3 10
16
One light-year 9.46 3 10
15
Mean orbit radius of the Earth 1.50 3 10
11
Mean distance from the Earth to the Moon 3.84 3 10
8
Distance from the equator to the North Pole 1.00 3 10
7
Mean radius of the Earth 6.37 3 10
6
Typical altitude (above the surface) of a satellite orbiting the Earth 2 3 10
5
Length of a football  eld 9.1 3 10
1
Length of this textbook 2.8 3 10
21
Length of a house y 5 3 10
23
Size of smallest visible dust particles , 10
24
Size of cells of most living organisms , 10
25
Diameter of a hydrogen atom , 10

210
Diameter of a uranium nucleus , 10
214
Diameter of a proton , 10
215

TABLE 1.2
|
Masses of Various Objects
(Approximate Values)
Mass (kg)
Visible , 10
52
Universe
Milky Way , 10
42
galaxy
Sun 1.99 3 10
30
Earth 5.98 3 10
24
Moon 7.36 3 10
22
Shark , 10
3
Human , 10
2
Frog , 10
21
Mosquito , 10

25
Bacterium , 10
215
Hydrogen 1.67 3 10
227
atom
Electron 9.11 3 10
231
c
De nition of the second
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1.2
|
Dimensional Analysis7
which is a ratio of mass to a product of three lengths. For example, aluminum has a
density of 2.70 3 10
3
kg/m
3
, and lead has a density of 11.3 3 10
3
kg/m
3
. An extreme
difference in density can be imagined by thinking about holding a 10-centimeter
(cm) cube of Styrofoam in one hand and a 10-cm cube of lead in the other.

1.2
|
Dimensional Analysis
In physics, the word dimension denotes the physical nature of a quantity. The dis-
tance between two points, for example, can be measured in feet, meters, or furlongs,
which are all different ways of expressing the dimension of length.
The symbols used in this book to specify the dimensions
2
of length, mass, and
time are L, M, and T, respectively. We shall often use square brackets [ ] to denote
the dimensions of a physical quantity. For example, in this notation the dimensions
of speed v are written [v] 5 L/T, and the dimensions of area A are [A] 5 L
2
. The di-
mensions of area, volume, speed, and acceleration are listed in Table 1.5, along with
their units in the two common systems. The dimensions of other quantities, such as
force and energy, will be described as they are introduced in the text.
In many situations, you may be faced with having to derive or check a speci c
equation. Although you may have forgotten the details of the derivation, a useful and
powerful procedure called dimensional analysis can be used as a consistency check, to
assist in the derivation, or to check your  nal expression. Dimensional analysis makes
use of the fact that dimensions can be treated as algebraic quantities. For example,
quantities can be added or subtracted only if they have the same dimensions. Fur-
thermore, the terms on both sides of an equation must have the same dimensions.
By following these simple rules, you can use dimensional analysis to help determine
TABLE 1.4
|
Some Prefi xes for
Powers of Ten
Power Pre x Abbreviation

10
224
yocto y
10
221
zepto z
10
218
atto a
10
215
femto f
10
212
pico p
10
29
nano n
10
26
micro ␮
10
23
milli m
10
22
centi c
10
21
deci d

10
3
kilo k
10
6
mega M
10
9
giga G
10
12
tera T
10
15
peta P
10
18
exa E
10
21
zetta Z
10
24
yotta Y
TABLE 1.3
|
Approximate Values of Some Time Intervals
Time
Interval (s)
Age of the Universe 4 3 10

17
Age of the Earth 1.3 3 10
17
Time interval since the fall of the Roman empire 5 3 10
12
Average age of a college student 6.3 3 10
8
One year 3.2 3 10
7
One day (time interval for one revolution of the
Earth about its axis)
8.6 3 10
4
One class period 3.0 3 10
3
Time interval between normal heartbeats 8 3 10
21
Period of audible sound waves , 10
23
Period of typical radio waves , 10
26
Period of vibration of an atom in a solid , 10
213
Period of visible light waves , 10
215
Duration of a nuclear collision , 10
222
Time interval for light to cross a proton , 10
224
Pitfall Prevention | 1.2

Symbols for Quantities
Some quantities have a small
number of symbols that represent
them. For example, the symbol
for time is almost always t. Other
quantities might have various symbols
depending on the usage. Length
may be described with symbols such
as x, y, and z (for position); r (for
radius); a, b, and c (for the legs of a
right triangle); ℓ (for the length of
an object); d (for a distance); h (for a
height); and so forth.
2
The dimensions of a variable will be symbolized by a capitalized, nonitalic letter, such as, in the case of length, L. The
symbol for the variable itself will be italicized, such as L for the length of an object or t for time.
TABLE 1.5
|
Dimensions and Units of Four Derived Quantities
Quantity Area (A) Volume (V ) Speed (v) Acceleration (a)
Dimensions L
2
L
3
L/T L/T
2
SI units m
2
m
3

m/s m/s
2
U.S. customary units ft
2
ft
3
ft/s ft/s
2
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8CHAPTER 1
|
Introduction and Vectors
whether an expression has the correct form because the relationship can be correct
only if the dimensions on the two sides of the equation are the same.
To illustrate this procedure, suppose you wish to derive an expression for the posi-
tion x of a car at a time t if the car starts from rest at t 5 0 and moves with constant
acceleration a. In Chapter 2, we shall  nd that the correct expression for this special
case is x 5
1
2
at
2
. Let us check the validity of this expression from a dimensional analy-
sis approach.
The quantity x on the left side has the dimension of length. For the equation to be
dimensionally correct, the quantity on the right side must also have the dimension of

length. We can perform a dimensional check by substituting the basic dimensions for
acceleration, L/T
2
(Table 1.5), and time, T, into the equation x 5
1
2
at
2
. That is, the
dimensional form of the equation x 5
1
2
at
2
can be written as
[x] 5
L
T
2
T
2
5 L
The dimensions of time cancel as shown, leaving the dimension of length, which is
the correct dimension for the position x. Notice that the number
1
2
in the equation
has no units, so it does not enter into the dimensional analysis.
QUICK QUIZ 1.1 True or False: Dimensional analysis can give you the numerical
value of constants of proportionality that may appear in an algebraic expression.

1.3
|
Conversion of Units
Sometimes it is necessary to convert units from one system to another or to convert
within a system, for example, from kilometers to meters. Equalities between SI and
U.S. customary units of length are as follows:

1
mi
l
e
(
mi
)
5
1
609 m 5
1
.609
k
m

1
f
t
5
0.30
4
8
m 5

30.
4
8

c
m

1 m 5 39.37 in. 5 3.281 f
t

1 inch (in.) 5 0.02
5
4 m 5 2.
5
4 cm
A more complete list of equalities can be found in Appendix A.
Units can be treated as algebraic quantities that can cancel each other. To per-
form a conversion, a quantity can be multiplied by a conversion factor, which is a frac-
tion equal to 1, with numerator and denominator having different units, to provide
the desired units in the  nal result. For example, suppose we wish to convert 15.0 in.
to centimeters. Because 1 in. 5 2.54 cm, we multiply by a conversion factor that is the
appropriate ratio of these equal quantities and  nd that
15.0 in. 5 (15.0 in.)
1
2.54 cm
1 in.
2
5 38.1 cm
Pitfall Prevention | 1.3
Always Include Units

When performing calculations, make
it a habit to include the units with
every quantity and carry the units
through the entire calculation. Avoid
the temptation to drop the units
during the calculation steps and
then apply the expected unit to the
number that results for an answer. By
including the units in every step, you
can detect errors if the units for the
answer are incorrect.
Example 1.1
|
Analysis of an Equation
Show that the expression v 5 at, where v represents speed, a acceleration, and t an instant of time, is dimensionally
correct.
SOLUTION
Identify the dimensions of v from Table 1.5: [v] 5
L
T
Identify the dimensions of a from Table 1.5 and multiply [at] 5
L
T
2
T 5
L
T
by the dimensions of t:
Therefore, v 5 at is dimensionally correct because we have the same dimensions on both sides. (If the expression were given
as v 5 at

2
, it would be dimensionally incorrect. Try it and see!)
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1.4
|
Order-of-Magnitude Calculations9
1.4
|
Order-of-Magnitude Calculations
Suppose someone asks you the number of bits of data on a typical musical compact
disc. In response, it is not generally expected that you would provide the exact num-
ber but rather an estimate, which may be expressed in scienti c notation. The esti-
mate may be made even more approximate by expressing it as an order of magnitude,
which is a power of ten determined as follows:
1. Express the number in scienti c notation, with the multiplier of the power of
ten between 1 and 10 and a unit.
2. If the multiplier is less than 3.162 (the square root of ten), the order of magni-
tude of the number is the power of ten in the scienti c notation. If the multi-
plier is greater than 3.162, the order of magnitude is one larger than the power
of ten in the scienti c notation.
We use the symbol , for “is on the order of.” Use the procedure above to verify
the orders of magnitude for the following lengths:
0.008 6 m , 10
22
m 0.002 1 m , 10
23

m 720 m , 10
3
m
Usually, when an order-of-magnitude estimate is made, the results are reliable to
within about a factor of ten. If a quantity increases in value by three orders of magni-
tude, its value increases by a factor of about 10
3
5 1 000.
where the ratio in parentheses is equal to 1. Notice that we express 1 as 2.54 cm/1 in.
(rather than 1 in./2.54 cm) so that the inch cancels with the unit in the original
quantity. The remaining unit is the centimeter, which is our desired result.
QUICK QUIZ 1.2 The distance between two cities is 100 mi. What is the number
of kilometers between the two cities? (a) smaller than 100 (b) larger than 100
(c) equal to 100
Example 1.2
|
Is He Speeding?
On an interstate highway in a rural region of Wyoming, a car is traveling at a speed of 38.0 m/s. Is the driver exceeding
the speed limit of 75.0 mi/h?
SOLUTION
Convert meters in the (38.0 m/s)
1
1 mi
1 609 m
2
5 2.36 3 10
22
mi/s
speed to miles:
Convert seconds to hours: (2.36 3 10

22
mi/s)
1
60 s
1 min
2

1
60 min
1 h

2
5 85.0 mi/h
The driver is indeed exceeding the speed limit and should slow down.
What If? What if the driver were from outside the United States and is
familiar with speeds measured in kilometers per hour? What is the speed of the
car in km/h?
Answer We can convert our  nal answer to the appropriate units:
(85.0 mi
/h)
1
1.609 km
1 mi
2
5 137 km/h
Figure 1.2 shows an automobile speedometer displaying speeds in both mi/h and
km/h. Can you check the conversion we just performed using this photograph?
Figure 1.2 (Example 1.2) The speedometer
of a vehicle that shows speeds in both miles
per hour and kilometers per hour.

© Cengage Learning/Ed Dodd
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10CHAPTER 1
|
Introduction and Vectors
Example 1.3
|
The Number of Atoms in a Solid
Estimate the number of atoms in 1 cm
3
of a solid.
SOLUTION
From Table 1.1 we note that the diameter d of an atom is about 10
210
m. Let us assume that the atoms in the solid are
spheres of this diameter. Then the volume of each sphere is about 10
230
m
3
(more precisely, volume 5 4␲ r
3
/3 5 ␲d
3
/6,
where r 5 d/2). Therefore, because 1 cm
3

5 10
26
m
3
, the number of atoms in the solid is on the order of 10
26
/10
230
5
10
24
atoms.
A more precise calculation would require additional knowledge that we could  nd in tables. Our estimate, however,
agrees with the more precise calculation to within a factor of 10.
Example 1.4
|
Breaths in a Lifetime
Estimate the number of breaths taken during an average human lifetime.
SOLUTION
We start by guessing that the typical human lifetime is about 70 years. Think about the average number of breaths that a
person takes in 1 min. This number varies depending on whether the person is exercising, sleeping, angry, serene, and so
forth. To the nearest order of magnitude, we shall choose 10 breaths per minute as our estimate. (This estimate is certainly
closer to the true average value than an estimate of 1 breath per minute or 100 breaths per minute.)
Find the approximate number of minutes in a year: 1 yr
1
400 days
1
y
r
21

25 h
1 da
y
21
60 min
1 h
2
5 6 3 10
5
min
Find the approximate number of minutes in a 70-year number of minutes 5 (70 yr)(6 3 10
5
min/yr)
lifetime: 5 4 3 10
7
min
Find the approximate number of breaths in a lifetime: number of breaths 5 (10 breaths/min)(4 3 10
7
min)
5 4 3 10
8
breaths
Therefore, a person takes on the order of 10
9
breaths in a lifetime. Notice how much simpler it is in the  rst calculation
above to multiply 400 3 25 than it is to work with the more accurate 365 3 24.
What If? What if the average lifetime were estimated as 80 years instead of 70? Would that change our  nal estimate?
Answer We could claim that (80 yr)(6 3 10
5
min/yr) 5 5 3 10

7
min, so our  nal estimate should be 5 3 10
8
breaths.
This answer is still on the order of 10
9
breaths, so an order-of-magnitude estimate would be unchanged.
1.5
|
Signifi cant Figures
When certain quantities are measured, the measured values are known only to within
the limits of the experimental uncertainty. The value of this uncertainty can depend
on various factors, such as the quality of the apparatus, the skill of the experimenter,
and the number of measurements performed. The number of signi cant  gures in a
measurement can be used to express something about the uncertainty. The number
of signi cant  gures is related to the number of numerical digits used to express the
measurement, as we discuss below.
As an example of signi cant  gures, suppose we are asked to measure the radius
of a compact disc using a meterstick as a measuring instrument. Let us assume the
accuracy to which we can measure the radius of the disc is 60.1 cm. Because of the
uncertainty of 60.1 cm, if the radius is measured to be 6.0 cm, we can claim only
that its radius lies somewhere between 5.9 cm and 6.1 cm. In this case, we say that the
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1.5
|
Signifi cant Figures11

measured value of 6.0 cm has two signi cant  gures. Note that the signi cant  gures
include the  rst estimated digit. Therefore, we could write the radius as (6.0 6 0.1) cm.
Zeros may or may not be signi cant  gures. Those used to position the decimal
point in such numbers as 0.03 and 0.007 5 are not signi cant. Therefore, there are
one and two signi cant  gures, respectively, in these two values. When the zeros
come after other digits, however, there is the possibility of misinterpretation. For
example, suppose the mass of an object is given as 1 500 g. This value is ambiguous
because we do not know whether the last two zeros are being used to locate the deci-
mal point or whether they represent signi cant  gures in the measurement. To re-
move this ambiguity, it is common to use scienti c notation to indicate the number
of signi cant  gures. In this case, we would express the mass as 1.5 3 10
3
g if there
are two signi cant  gures in the measured value, 1.50 3 10
3
g if there are three sig-
ni cant  gures, and 1.500 3 10
3
g if there are four. The same rule holds for numbers
less than 1, so 2.3 3 10
24
has two signi cant  gures (and therefore could be written
0.000 23) and 2.30 3 10
24
has three signi cant  gures (also written as 0.000 230).
In problem solving, we often combine quantities mathematically through mul-
tiplication, division, addition, subtraction, and so forth. When doing so, you must
make sure that the result has the appropriate number of signi cant  gures. A good
rule of thumb to use in determining the number of signi cant  gures that can be
claimed in a multiplication or a division is as follows:

When multiplying several quantities, the number of signi cant  gures in the  nal
answer is the same as the number of signi cant  gures in the quantity having the
smallest number of signi cant  gures. The same rule applies to division.
Let’s apply this rule to  nd the area of the compact disc whose radius we mea-
sured above. Using the equation for the area of a circle,
A 5 ␲r
2
5 ␲(6.0 cm)
2
5 1.1 3 10
2
cm
2
If you perform this calculation on your calculator, you will likely see 113.097 335 5.
It should be clear that you don’t want to keep all of these digits, but you might be
tempted to report the result as 113 cm
2
. This result is not justi ed because it has
three signi cant  gures, whereas the radius only has two. Therefore, we must report
the result with only two signi cant  gures as shown above.
For addition and subtraction, you must consider the number of decimal places
when you are determining how many signi cant  gures to report:
When numbers are added or subtracted, the number of decimal places in the re-
sult should equal the smallest number of decimal places of any term in the sum
or difference.
As an example of this rule, consider the sum
23.2 1 5.174 5 28.4
Notice that we do not report the answer as 28.374 because the lowest number of deci-
mal places is one, for 23.2. Therefore, our answer must have only one decimal place.
The rules for addition and subtraction can often result in answers that have a

different number of signi cant  gures than the quantities with which you start. For
example, consider these operations that satisfy the rule:
1.000 1 1 0.000 3 5 1.000 4
1.002 2 0.998 5 0.004
In the  rst example, the result has  ve signi cant  gures even though one of the
terms, 0.000 3, has only one signi cant  gure. Similarly, in the second calculation,
the result has only one signi cant  gure even though the numbers being subtracted
have four and three, respectively.
Pitfall Prevention | 1.4
Read Carefully
Notice that the rule for addition
and subtraction is different from
that for multiplication and division.
For addition and subtraction, the
important consideration is the
number of decimal places, not the
number of signi cant  gures.
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12CHAPTER 1
|
Introduction and Vectors
In this book, most of the numerical examples and end-of-chapter problems
will yield answers having three signi cant  gures. When carrying out estima-
tion calculations, we shall typically work with a single signi cant  gure.
If the number of signi cant  gures in the result of a calculation must be reduced,
there is a general rule for rounding numbers: the last digit retained is increased by 1

if the last digit dropped is greater than 5. (For example, 1.346 becomes 1.35.) If the
last digit dropped is less than 5, the last digit retained remains as it is. (For example,
1.343 becomes 1.34.) If the last digit dropped is equal to 5, the remaining digit
should be rounded to the nearest even number. (This rule helps avoid accumulation
of errors in long arithmetic processes.)
A technique for avoiding error accumulation is to delay the rounding of numbers
in a long calculation until you have the  nal result. Wait until you are ready to copy the
 nal answer from your calculator before rounding to the correct number of signi cant
 gures. In this book, we display numerical values rounded off to two or three signi -
cant  gures. This occasionally makes some mathematical manipulations look odd or
incorrect. For instance, looking ahead to Example 1.8 on page 21, you will see the op-
eration 217.7 km 1 34.6 km 5 17.0 km. This looks like an incorrect subtraction, but
that is only because we have rounded the numbers 17.7 km and 34.6 km for display.
If all digits in these two intermediate numbers are retained and the rounding is only
performed on the  nal number, the correct three-digit result of 17.0 km is obtained.
c
Signi cant  gure guidelines
used in this book
Pitfall Prevention | 1.5
Symbolic Solutions
When solving problems, it is very
useful to perform the solution
completely in algebraic form and wait
until the very end to enter numerical
values into the  nal symbolic
expression. This method will save
many calculator keystrokes, especially
if some quantities cancel so that you
never have to enter their values into
your calculator! In addition, you will

only need to round once, on the
 nal result.
1.6
|
Coordinate Systems
Many aspects of physics deal in some way or another with locations in space. For ex-
ample, the mathematical description of the motion of an object requires a method
for specifying the object’s position. Therefore, we  rst discuss how to describe the
position of a point in space by means of coordinates in a graphical representation.
A point on a line can be located with one coordinate, a point in a plane is located
with two coordinates, and three coordinates are required to locate a point in space.
A coordinate system used to specify locations in space consists of
• A  xed reference point O, called the origin
• A set of speci ed axes or directions with an appropriate scale and labels on the axes
• Instructions that tell us how to label a point in space relative to the origin and axes
One convenient coordinate system that we will use frequently is the Cartesian co-
ordinate system, sometimes called the rectangular coordinate system. Such a system in two
dimensions is illustrated in Figure 1.3. An arbitrary point in this system is labeled with
the coordinates (x, y). Positive x is taken to the right of the origin, and positive y is
upward from the origin. Negative x is to the left of the origin, and negative y is down-
ward from the origin. For example, the point P, which has coordinates (5, 3), may be
reached by going  rst 5 m to the right of the origin and then 3 m above the origin
Example 1.5
|
Installing a Carpet
A carpet is to be installed in a rectangular room whose length is measured to be 12.71 m and whose width is measured to
be 3.46 m. Find the area of the room.
SOLUTION
If you multiply 12.71 m by 3.46 m on your calculator, you will see an answer of 43.976 6 m
2

. How many of these numbers
should you claim? Our rule of thumb for multiplication tells us that you can claim only the number of signi cant  gures in
your answer as are present in the measured quantity having the lowest number of signi cant  gures. In this example, the
lowest number of signi cant  gures is three in 3.46 m, so we should express our  nal answer as 44.0 m
2
.
Figure 1.3 Designation of points
in a Cartesian coordinate system.
Each square in the xy plane is 1 m
on a side. Every point is labeled with
coordinates (x, y).
y
x
Q
(–3, 4)
(5, 3)
(x, y)
P
5
10
5
10
O
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1.7
|

Vectors and Scalars13
(or by going 3 m above the origin and then 5 m to the right). Similarly, the point Q
has coordinates (23, 4), which correspond to going 3 m to the left of the origin and
4 m above the origin.
Sometimes it is more convenient to represent a point in a plane by its plane polar
coordinates (r, ␪), as in Active Figure 1.4a. In this coordinate system, r is the length
of the line from the origin to the point, and ␪ is the angle between that line and a
 xed axis, usually the positive x axis, with ␪ measured counterclockwise. From the
right triangle in Active Figure 1.4b, we  nd that sin ␪ 5 y/r and cos ␪ 5 x/r. (A
review of trigonometric functions is given in Appendix B.4.) Therefore, starting
with plane polar coordinates, one can obtain the Cartesian coordinates through
the equations

x 5 r cos
␪
1.2
b

y
5 r sin ␪ 1.3
b
Furthermore, if we know the Cartesian coordinates, the de nitions of trigonometry tell
us that
tan ␪ 5
y
x
1.4
b
and
r 5

"
x
2
1
y

2
1.5
b
You should note that these expressions relating the coordinates (x, y) to the coor-
dinates (r, ␪) apply only when ␪ is de ned as in Active Figure 1.4a, where positive ␪ is
an angle measured counterclockwise from the positive x axis. Other choices are made in
navigation and astronomy. If the reference axis for the polar angle ␪ is chosen to be
other than the positive x axis or if the sense of increasing ␪ is chosen differently, the
corresponding expressions relating the two sets of coordinates will change.
1.7
|
Vectors and Scalars
Each of the physical quantities that we shall encounter in this text can be placed in
one of two categories, either a scalar or a vector. A scalar is a quantity that is com-
pletely speci ed by a positive or negative number with appropriate units. On the
other hand, a vector is a physical quantity that must be speci ed by both magnitude
and direction.
The number of grapes in a bunch (Fig. 1.5a) is an example of a scalar quantity. If
you are told that there are 38 grapes in the bunch, this statement completely speci es
Active Figure 1.4 (a) The plane
polar coordinates of a point are
represented by the distance r and
the angle ␪, where ␪ is measured in a
counterclockwise direction from the

positive x axis. (b) The right triangle
used to relate (x, y) to (r, ␪).
O
(x, y)
y
x
r
x
r
y
sin =
y
r
cos =
x
r
tan =
x
y
u
u
u
u
u
a
b
Figure 1.5 (a) The number of grapes in this bunch is one example of a scalar quantity.
Can you think of other examples? (b) This helpful person pointing in the correct direction
tells us to travel  ve blocks north to reach the courthouse. A vector is a physical quantity
that is speci ed by both magnitude and direction.

a
b
Mack Henley/Visuals Unlimited, Inc.
© Cengage Learning/George Semple
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Licensed to:
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14CHAPTER 1
|
Introduction and Vectors
the information; no speci cation of direction is required. Other examples of
scalars are temperature, volume, mass, and time intervals. The rules of ordi-
nary arithmetic are used to manipulate scalar quantities; they can be freely
added and subtracted (assuming that they have the same units!), multiplied
and divided.
Force is an example of a vector quantity. To describe the force on an object com-
pletely, we must specify both the direction of the applied force and the magnitude of
the force.
Another simple example of a vector quantity is the displacement of a particle, de-
 ned as its change in position. The person in Figure 1.5b is pointing out the direction
of your desired displacement vector if you would like to reach a destination such as
the courthouse. She will also tell you the magnitude of the displacement along with
the direction, for example, “5 blocks north.”
Suppose a particle moves from some point 𝖠 to a point 𝖡 along a straight
path, as in Figure 1.6. This displacement can be represented by drawing an
arrow from 𝖠 to 𝖡, where the arrowhead represents the direction of the dis-
placement and the length of the arrow represents the magnitude of the displace-
ment. If the particle travels along some other path from 𝖠 to 𝖡, such as the

broken line in Figure 1.6, its displacement is still the vector from 𝖠 to 𝖡. The
vector displacement along any indirect path from 𝖠 to 𝖡 is defined as being
equivalent to the displacement represented by the direct path from 𝖠 to 𝖡. The
magnitude of the displacement is the shortest distance between the end points.
Therefore, the displacement of a particle is completely known if its initial and
final coordinates are known. The path need not be specified. In other words,
the displacement is independent of the path if the end points of the path
are fixed.
Note that the distance traveled by a particle is distinctly different from its displace-
ment. The distance traveled (a scalar quantity) is the length of the path, which in
general can be much greater than the magnitude of the displacement. In Figure 1.6,
the length of the curved broken path is much larger than the magnitude of the solid
black displacement vector.
If the particle moves along the x axis from position x
i
to position x
f
, as in Fig-
ure1.7, its displacement is given by x
f
2 x
i
. (The indices i and f refer to the initial
and  nal values.) We use the Greek letter delta (D) to denote the change in a quan-
tity. Therefore, we de ne the change in the position of the particle (the displace-
ment) as

Dx ; x
f
2 x

i
1.6
b
From this de nition we see that Dx is positive if x
f
is greater than x
i
and negative if
x
f
is less than x
i
. For example, if a particle changes its position from x
i
5 2 5 m to
x
f
5 3 m, its displacement is Dx 5 1 8 m.
Many physical quantities in addition to displacement are vectors. They include
velocity, acceleration, force, and momentum, all of which will be de ned in later
chapters. In this text, we will use boldface letters with an arrow over the letter, suchas
A
:
, to represent vectors. Another common notation for vectors with which you should
be familiar is a simple boldface character: A.
The magnitude of the vector
A
:
is written with an italic letter A or, alternatively,
u

A
:
u
. The magnitude of a vector is always positive and carries the units of the quantity
that the vector represents, such as meters for displacement or meters per second
for velocity. Vectors combine according to special rules, which will be discussed in
Sections 1.8 and 1.9.
QUICK QUIZ 1.3 Which of the following are vector quantities and which are
scalar quantities? (a) your age (b) acceleration (c) velocity (d) speed
(e) mass
c
Displacement
c
Distance
Figure 1.6 After a particle moves
from 𝖠 to 𝖡 along an arbitrary
path represented by the broken
line, its displacement is a vector
quantity shown by the arrow drawn
from 𝖠 to Ꭾ.
Ꭽ
Ꭾ
Figure 1.7 A particle moving along
the x axis from x
i
to x
f
undergoes a
displacement Dx 5 x
f

2 x
i
.
O
y
x
x
f
x
i
∆x
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Licensed to:
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1.8
|
Some Properties of Vectors15
1.8
|
Some Properties of Vectors
Equality of Two Vectors
Two vectors
A
:
and
B
:
are de ned to be equal if they have the same units, the same

magnitude, and the same direction. That is,
A
:
5
B
:
only if A 5 B and
A
:
and
B
:
point
in the same direction. For example, all the vectors in Figure 1.8 are equal even
though they have different starting points. This property allows us to translate a vec-
tor parallel to itself in a diagram without affecting the vector.
Addition
The rules for vector sums are conveniently described using a graphical method.
To add vector
B
:
to vector
A
:
,  rst draw a diagram of vector
A
:
on graph paper, with its
mag nitude represented by a convenient scale, and then draw vector
B

:
to the same scale
with its tail starting from the tip of
A
:
, as in Active Figure 1.9a. The resultant vector
R
:
5
A
:
1
B
:
is the vector drawn from the tail of
A
:
to the tip of
B
:
. The technique for
adding two vectors is often called the “head-to-tail method.”
When vectors are added, the sum is independent of the order of the addition. This
independence can be seen for two vectors from the geometric construction in Active
Figure 1.9b and is known as the commutative law of addition:

A
:
1
B

:
5
B
:
1
A
:
1.7
b
Figure 1.8 These four represent a-
tions of vectors are equal because all
four vectors have the same magnitude
and point in the same direction.
O
y
x
THINKING PHYSICS 1.1
Consider your commute to work or school in the morning. Which is larger, the
distance you travel or the magnitude of the displacement vector?
Reasoning Unless you have a very unusual commute, the distance traveled must
be larger than the magnitude of the displacement vector. The distance includes
the results of all the twists and turns you make in following the roads from home
to work or school. On the other hand, the magnitude of the displacement vector
is the length of a straight line from your home to work or school. This length is
often described informally as “the distance as the crow  ies.” The only way that
the distance could be the same as the magnitude of the displacement vector is
if your commute is a perfect straight line, which is highly unlikely! The distance
could never be less than the magnitude of the displacement vector because the
shortest distance between two points is a straight line.
b

ϭ
ϩ
A
S

R
S
A
S
B
S
B
S
a
A
S
B
S
B
S
A
S
A
S
B
S
B
S
R
S

ϭϭϩϩ
Draw ,
then add .
A
S
B
S
A
S
Draw ,
then add .
A
S
B
S
b
Active Figure 1.9 (a) When vector
B
:
is added to vector A
:
,
the resultant
R
:
is the
vector that runs from the tail of
A
:
to the tip of

B
:
. (b) This construction shows that
A
:
1
B
:
5
B
:
1 A
:
; vector addition is commutative.
Pitfall Prevention | 1.6
Vector Addition Versus Scalar
Addition
Keep in mind that
A
:
1
B
:
5
C
:
is very
different from A 1 B 5 C. The  rst
equation is a vector sum, which must
be handled carefully, such as with the

graphical method described in Active
Figure 1.9. The second equation is a
simple algebraic addition of numbers
that is handled with the normal rules
of arithmetic.
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Licensed to:
CengageBrain User
16CHAPTER 1
|
Introduction and Vectors
If three or more vectors are added, their sum is independent of the way in which
they are grouped. A geometric demonstration of this property for three vectors is
given in Figure 1.10. This property is called the associative law of addition:

A
:
1 (B
:
1 C
:
) 5 (A
:
1 B
:
) 1 C
:
1.8

b
Geometric constructions can also be used to add more than three vectors,
as shown in Figure 1.11 for the case of four vectors. The resultant vector
R
:
5
A
:
1
B
:
1
C
:
1
D
:
is the vector that closes the polygon formed by the vectors being added.
In other words,
R
:
is the vector drawn from the tail of the  rst vector to the tip of the
last vector. Again, the order of the summation is unimportant.
In summary, a vector quantity has both magnitude and direction and also
obeysthe laws of vector addition as described in Active Figure 1.9 and Figures 1.10
and 1.11. When two or more vectors are added together, they must all have the same
units and they must all be the same type of quantity. It would be meaningless to add
a velocity vector (for example, 60 km/h to the east) to a displacement vector (for
example, 200 km to the north) because these vectors represent different physical
quantities. The same rule also applies to scalars. For example, it would be meaning-

less to add time intervals to temperatures.
Negative of a Vector
The negative of the vector
A
:
is de ned as the vector that, when added to
A
:
, gives
zero for the vector sum. That is,
A
:
1 (2
A
:
) 5 0. The vectors
A
:
and 2
A
:
have the same
magnitude but point in opposite directions.
Subtraction of Vectors
The operation of vector subtraction makes use of the de nition of the negative of a
vector. We de ne the operation
A
:
2
B

:
as vector 2
B
:
added to vector
A
:
:

A
:
2 B
:
5 A
:
1 (2B
:
) 1.9
b
The geometric construction for subtracting two vectors is illustrated in Figure 1.12.
Multiplication of a Vector by a Scalar
If a vector
A
:
is multiplied by a positive scalar quantity s, the product s
A
:
is a vector
that has the same direction as
A

:
and magnitude sA. If s is a negative scalar quantity, the
vector s
A
:
is directed opposite to
A
:
. For example, the vector 5
A
:
is  ve times longer
than
A
:
and has the same direction as
A
:
; the vector 2
1
3
A
:
has one-third the magni-
tude of
A
:
and points in the direction opposite
A
:

.
Multiplication of Two Vectors
Two vectors
A
:
and
B
:
can be multiplied in two different ways to produce either a sca-
lar or a vector quantity. The scalar product (or dot product)
A
:
?
B
:
is a scalar quantity
equal to AB cos ␪, where ␪ is the angle between
A
:
and
B
:
. The vector product (or cross
product)
A
:
3
B
:
is a vector quantity whose magnitude is equal to AB sin ␪. We shall

discuss these products more fully in Chapters 6 and 10, where they are  rst used.
QUICK QUIZ 1.4 The magnitudes of two vectors
A
:
and
B
:
are A 5 12 units and
B 5 8 units. Which pair of numbers represents the largest and smallest possible values
for the magnitude of the resultant vector
R
:
5
A
:
1
B
:
? (a) 14.4 units, 4 units
(b) 12 units, 8 units (c) 20 units, 4 units (d) none of these answers
QUICK QUIZ 1.5 If vector
B
:
is added to vector
A
:
, under what condition does the
resultant vector
A
:

1
B
:
have magnitude A 1 B? (a)
A
:
and
B
:
are parallel and in the
same direction. (b)
A
:
and
B
:
are parallel and in opposite directions. (c)
A
:
and
B
:
are
perpendicular.
Figure 1.10 Geometric constructions
for verifying the associative law of
addition.
Add and ;
then add the
result to .

A
S

A
S

A
S

A
S

B
S

B
S

B
S

B
S

B
S

ϩ
C
S

C
S

C
S
Add and ;
then add to
the result.
A
S

B
S

C
S
ϩ
C
S

A
S
B
S

ϩϩ
C
S

)(
A
S

B
S

ϩϩ
C
S

)
(
Figure 1.11 Geometric construction
for summing four vectors. The
resultant vector
R
:
closes the polygon
and points from the tail of the  rst
vector to the tip of the  nal vector.
A
S
B
S
C
S
D
S
A

S
B
S
C
S
D
S
R
S
ϭ
ϩ
ϩ
ϩ
Figure 1.12 Subtracting vector
B
:
from vector A
:
. The vector 2
B
:
is
equal in magnitude to vector
B
:
and
points in the opposite direction.
Ϫ
A
S

A
S
B
S
B
S
B
S
Ϫ
A
S
B
S
We would draw
here if we were
adding it to .
Adding Ϫ to
is equivalent to
subtracting
from .
B
S
A
S
A
S
B
S
10293_ch01_001-034.indd 1610293_ch01_001-034.indd 16 1/6/12 2:35 PM1/6/12 2:35 PM
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1.9
|
Components of a Vector and Unit Vectors17
1.9
|
Components of a Vector and Unit Vectors
The graphical method of adding vectors is not recommended whenever high
accuracy is required or in three-dimensional problems. In this section, we de-
scribe a method of adding vectors that makes use of the projections of vectors
along coordinate axes. These projections are called the components of the vec-
tor or its rectangular components. Any vector can be completely described by its
components.
Consider a vector
A
:
lying in the xy plane and making an arbitrary angle ␪
with the positive x axis as shown in Figure 1.13a. This vector can be expressed as
the sum of two other component vectors A
:
x
, which is parallel to the x axis, and
A
:
y
, which
is parallel to the y axis. From Figure 1.13b, we see that the three vectors form a right
triangle and that A

:
5 A
x
:
1 A
y
:
. We shall often refer to the “components of a vector
A
:
,” written A
x
and A
y
(without the boldface notation). The component A
x
rep-
resents the projection of
A
:
along the x axis, and the component A
y
represents
the projection of
A
:
along the y axis. These components can be positive or negative.
The component A
x
is positive if the component vector

A
:
x
points in the positive x
direction and is negative if
A
:
x
points in the negative x direction. A similar statement
is made for the component A
y
.
From Figure 1.13b and the de nition of the sine and cosine of an angle, we see
that cos ␪ 5 A
x
/A and sin ␪ 5 A
y
/A. Hence, the components of
A
:
are given by
A
x
5 A cos ␪ and A
y
5 A sin ␪ 1.10
b
The magnitudes of these components are the lengths of the two sides of a right tri-
angle with a hypotenuse of length A. Therefore, the magnitude and direction of
A

:

are related to its components through the expressions
A 5
"
A
x

2
1 A
y

2
1.11
b
tan ␪ 5
A
y
A
x
1.12
b
To solve for ␪, we can write ␪ 5 tan
21
(A
y
/A
x
), which is read “␪ equals the angle whose
tangent is the ratio A

y
/A
x
.” Note that the signs of the components A
x
and A
y
depend on the
angle ␪. For example, if ␪ 5 1208, A
x
is negative and A
y
is positive. If ␪ 5 2258, both A
x
and A
y
are negative. Figure 1.14 summarizes the signs of the components when
A
:
lies in the various quadrants.
If you choose reference axes or an angle other than those shown in Figure 1.13,
the components of the vector must be modi ed accordingly. In many applications, it
is more convenient to express the components of a vector in a coordinate system hav-
ing axes that are not horizontal and vertical but are still perpendicular to each other.
c
Magnitude of
:
c
Direction of
:

Figure 1.14 The signs of the
components of a vector
A
:
depend on
the quadrant in which the vector is
located.
y
A
x
points
left and is Ϫ
A
x
points
right and is ϩ
A
y
points
up and is ϩ
A
y
points
up and is ϩ
A
x
points
left and is Ϫ
A
x

points
right and is ϩ
A
y
points
down and is Ϫ
A
y
points
down and is Ϫ
x
Pitfall Prevention | 1.7
x and y Components
Equation 1.10 associates the cosine
of the angle with the x component
and the sine of the angle with the
y component. This association is
true only because we measured the
angle ␪ with respect to the x axis, so
do not memorize these equations.
If ␪ is measured with respect to the
y axis (as in some problems), these
equations will be incorrect. Think
about which side of the triangle
containing the components is
adjacent to the angle and which side
is opposite and then assign the cosine
and sine accordingly.
Figure 1.13 (a) A vector
A

:
lying in the xy plane can be represented by
its component vectors A
:
x
and A
:
y
. (b) The y component vector A
:
y
can be
movedto the right so that it adds to A
:
x
. The vector sum of the component
vectors is A
:
. These three vectors form a right triangle.
y
x
O
y
x
y
x
O
x
y
u

u
A
S
A
S
A
S
A
S
A
S
A
S
a b
10293_ch01_001-034.indd 1710293_ch01_001-034.indd 17 1/6/12 2:35 PM1/6/12 2:35 PM
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18CHAPTER 1
|
Introduction and Vectors
Suppose a vector
B
:
makes an angle ␪9 with the x9 axis de ned in Figure 1.15. The com-
ponents of
B
:
along these axes are given by B

x9
5 B cos ␪9 and B
y9
5 B sin ␪9, as in Equa-
tion 1.10. The magnitude and direction of
B
:
are obtained from expressions equivalent
to Equations 1.11 and 1.12. Therefore, we can express the components of a vector in
any coordinate system that is convenient for a particular situation.
QUICK QUIZ 1.6 Choose the correct response to make the sentence true: A
component of a vector is (a) always, (b) never, or (c) sometimes larger than the
magnitude of the vector.
Unit Vectors
Vector quantities often are expressed in terms of unit vectors. A unit vector is a
dimensionless vector having a magnitude of exactly 1. Unit vectors are used to specify
a given direction and have no other physical signi cance. We shall use the symbols i
ˆ
,

j
ˆ
, and
k
ˆ
to represent unit vectors pointing in the x, y, and z directions, respectively.
The “hat” over the letters is a common notation for a unit vector; for example,
i
ˆ
is

called “i-hat.” The unit vectors i
ˆ
,

j
ˆ
, and
k
ˆ
form a set of mutually perpendicular vectors
as shown in Active Figure 1.16a, where the magnitude of each unit vector equals 1;
that is, u i
ˆ
u 5 u
j
ˆ
u 5 uk
ˆ
u 5 1.
Consider a vector
A
:
lying in the xy plane, as in Active Figure 1.16b. The product of
the component A
x
and the unit vector
i
ˆ
is the component vector A
:

x
5 A
x
i
ˆ
, which lies
on the x axis and has magnitude A
x
. Likewise, A
y
j
ˆ
is a component vector of magnitude
A
y
lying on the y axis. Therefore, the unit- vector notation for the vector
A
:
is

A
:
5 A
x
i
ˆ
1 A
y
j
ˆ

1.13
b
Now suppose we wish to add vector
B
:
to vector A
:
,
where
B
:
has components B
x
and
B
y
. The procedure for performing this sum is simply to add the x and y com ponents
separately. The resultant vector
R
:
5
A
:
1
B
:
is therefore

R
:

5 (A
x
1 B
x
)i
ˆ
1 (A
y
1 B
y
)j
ˆ
1.14
b
From this equation, the components of the resultant vector are given by
R
x
5 A
x
1 B
x

1.15
b

R
y
5 A
y
1 B

y
Therefore, we see that in the component method of adding vectors, we add all the
x components to  nd the x component of the resultant vector and use the same
Active Figure 1.16 (a) The unit
vectors i
ˆ
,
j
ˆ
, and
k
ˆ
are directed along
the x, y, and z axes, respectively.
(b) A vector
A
:
lying in the xy plane
has component vectors A
x
i
ˆ
and A
y
j
ˆ
,
where A
x
and A

y
are the components
of A
:
.
y
x
A
x
A
y
ˆ
j
ˆ
i
x
y
z
ˆ
j
ˆ
i
ˆ
k
a b
A
S
Figure 1.15 The component vectors
of vector
B

:
in a coordinate system
that is tilted.
x′
y′
′
B
y′
B
x′
O′
θ
B
S
10293_ch01_001-034.indd 1810293_ch01_001-034.indd 18 1/6/12 2:35 PM1/6/12 2:35 PM
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deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Licensed to:
CengageBrain User
1.9
|
Components of a Vector and Unit Vectors19
process for the y components. The procedure just described for adding two vec-
tors
A
:
and
B
:
using the component method can be checked using a diagram like

Figure1.17.
The magnitude of
R
:
and the angle it makes with the x axis can be obtained from
its components using the relationships

R 5
"
R
x

2
1 R
y

2
5
"
(A
x
1 B
x
)
2
1 (A
y
1 B
y
)

2
1.16
b

tan ␪ 5
R
y
R
x
5
A
y
1 B
y
A
x
1 B
x
1.17
b
The extension of these methods to three-dimensional vectors is straightforward. If
A
:
and
B
:
both have x, y, and z components, we express them in the form
A
:
5 A

x
i
ˆ
1 A
y
j
ˆ
1 A
z

k
ˆ
B
:
5 B
x
i
ˆ
1 B
y
j
ˆ
1 B
z

k
ˆ
The sum of
A
:

and
B
:
is

R
:
5 A
:
1 B
:
5 (A
x
1 B
x
)i
ˆ
1 (A
y
1 B
y

)j
ˆ
1 (A
z
1 B
z
)
k

ˆ
1.18
b
If a vector
R
:
has x, y, and z components, the magnitude of the vector is
R 5
"
R
x

2
1 R
y

2
1 R
z

2
The angle ␪
x
that
R
:
makes with the x axis is given by
cos ␪
x
5

R
x
R
with similar expressions for the angles with respect to the y and z axes.
The extension of our method to adding more than two vectors is also
straightforward. For example,
A
:
1
B
:
1
C
:
5 (A
x
1 B
x
1 C
x
)
i
ˆ
1 (A
y
1 B
y
1 C
y
)

j
ˆ
1
(A
z
1 B
z
1 C
z
)
k
ˆ
. Adding displacement vectors is relatively easy to visualize. We can
also add other types of vectors, such as velocity, force, and electric  eld vectors,
which we will do in later chapters.
QUICK QUIZ 1.7 If at least one component of a vector is a positive number, the
vector cannot (a) have any component that is negative, (b) be zero, (c) have three
dimensions.
QUICK QUIZ 1.8 If
A
:
1
B
:
5 0, the corresponding components of the two vectors
A
:
and
B
:

must be (a) equal, (b) positive, (c) negative, (d) of opposite sign.
Figure 1.17 A geometric
construction showing the relation
between the components of the
resultant
R
:
of two vectors and the
individual components.
y
x
B
x
A
y
A
x
R
x
B
y
R
y
A
S
B
S
R
S
Pitfall Prevention | 1.8

Tangents on Calculators
Equation 1.17 involves the calculation
of an angle by means of a tangent
function. Generally, the inverse
tangent function on calculators
provides an angle between 2908
and 1908. As a consequence, if the
vector you are studying lies in the
second or third quadrant, the angle
measured from the positive x axis will
be the angle your calculator returns
plus1808.
THINKING PHYSICS 1.2
You may have asked someone directions to a destination in a city and been told
something like, “Walk 3 blocks east and then 5 blocks south.” If so, are you expe-
rienced with vector components?
Reasoning Yes, you are! Although you may not have thought of vector compo-
nent language when you heard these directions, that is exactly what the direc-
tions represent. The perpendicular streets of the city re ect an xy coordinate
system; we can assign the x axis to the east–west streets, and the y axis to the
north–south streets. Therefore, the comment of the person giving you directions
can be translated as, “Undergo a displacement vector that has an x component
of 13 blocks and a y component of 25 blocks.” You would arrive at the same
destination by undergoing the y component  rst, followed by the x component,
demonstrating the commutative law of addition.
b
10293_ch01_001-034.indd 1910293_ch01_001-034.indd 19 1/6/12 2:35 PM1/6/12 2:35 PM
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deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Licensed to:

CengageBrain User
20CHAPTER 1
|
Introduction and Vectors
Example 1.6
|
The Sum of Two Vectors
Find the sum of two displacement vectors
A
:
and
B
:
lying in the xy plane and given by
A
:
5 (2.0
i
ˆ
1 2.0
j
ˆ
) m and
B
:
5 (2.0
i
ˆ
2 4.0
j

ˆ
) m
SOLUTION
Comparing this expression for
A
:
with the general expression
A
:
5 A
x
i
ˆ
1 A
y
j
ˆ
1 A
z
k
ˆ
, we see that A
x
5 2.0 m, A
y
5 2.0 m,
and A
z
5 0. Likewise, B
x

5 2.0 m, B
y
5 24.0 m, and B
z
5 0. We can use a two-dimensional approach because there are no
z components.
Use Equation 1.14 to obtain the resultant vector
R
:
:
R
:
5
A
:
1
B
:
5 (2.0 1 2.0)
i
ˆ
m 1 (2.0 2 4.0)
j
ˆ
m
Evaluate the components of
R
:
: R
x

5 4.0 m R
y
5 22.0 m
Use Equation 1.16 to  nd the magnitude of
R
:
: R 5
"
R
x

2
1 R
y

2
5
"
(4.0 m)
2
1 (22.0 m)
2
5
"
20 m 5 4.5 m
Find the direction of
R
:
from Equation 1.17: tan ␪ 5
R

y
R
x
5
22.0 m
4.0 m
520.50
Your calculator likely gives the answer 2278 for ␪ 5 tan
21
(20.50). This answer is correct if we interpret it to mean 278
clockwise from the x axis. Our standard form has been to quote the angles measured counterclockwise from the 1x axis,
and that angle for this vector is ␪ 5 333.8
Example 1.7
|
The Resultant Displacement
A particle undergoes three consecutive displacements: D
:
r
1
5 (15
i
ˆ
1 30
j
ˆ
1 12
k
ˆ
) cm, D
:

r
2
5 (23
i
ˆ
2 14
j
ˆ
2 5.0k
ˆ
) cm, and
D
:
r
3
5 (213
i
ˆ
1 15
j
ˆ
) cm. Find unit-vector notation for the resultant displacement and its magnitude.
SOLUTION
Although x is suf cient to locate a point in one dimension, we need a vector
:
r
to locate a point in two or three dimen-
sions. The notation
D
:

r
is a generalization of the one-dimensional displacement Dx. Three-dimensional displacements
are more dif cult to conceptualize than those in two dimensions because the latter can be drawn on paper.
For this problem, let us imagine that you start with your pencil at the origin of a piece of graph paper on which you
have drawn x and y axes. Move your pencil 15 cm to the right along the x axis, then 30 cm upward along the y axis, and then
12 cm perpendicularly toward you away from the graph paper. This procedure provides the displacement described by D
:
r
1
. From
this point, move your pencil 23cm to the right parallel to the x axis, then 14 cm parallel to the graph paper in the 2y direc-
tion, and then 5.0 cm perpendicularly away from you toward the graph paper. You are now at the displacement from the origin
described by D
:
r
1
1 D
:
r
2
. From this point, move your pencil 13 cm to the left in the 2x direction, and ( nally!) 15 cm parallel
to the graph paper along the y axis. Your  nal position is at a displacement D
:
r
1
1 D
:
r
2
1 D

:
r
3
from the origin.
To  nd the resultant displacement, D
:
r

5 D
:
r
1
1 D
:
r
2
1 D
:
r
3
add the three vectors:
5 (15 1 23 2 13)
i
ˆ
cm 1 (30 2 14 1 15)
j
ˆ
cm
1 (12 2 5.0 1 0)
k

ˆ
cm
5 (25
i
ˆ
1 31
j
ˆ
1 7.0
k
ˆ
) cm
Find the magnitude of the resultant vector: R 5 "R
2
x
1 R
2
y
1 R
2
z
5 "(25 cm)
2
1 (31 cm)
2
1 (7.0 cm)
2
5 40 cm
10293_ch01_001-034.indd 2010293_ch01_001-034.indd 20 1/6/12 2:35 PM1/6/12 2:35 PM
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deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Licensed to:
CengageBrain User
1.9
|
Components of a Vector and Unit Vectors21
Example 1.8
|
Taking a Hike
A hiker begins a trip by first walking 25.0 km southeast from her car. She stops
and sets up her tent for the night. On the second day, she walks 40.0 km in a di-
rection 60.08 north of east, at which point she discovers a forest ranger’s tower.
(A) Determine the components of the hiker’s displacement for each day.
SOLUTION
If we denote the displacement vectors on the  rst and second days by
A
:
and
B
:
, respectively, and use the car as the origin of coordinates, we obtain the vectors
shown in Figure 1.18. Drawing the resultant
R
:
, we see that this problem is one we’ve
solved before: an addition of two vectors.
Displacement
A
:
has a magnitude of 25.0 km and is directed 45.08 below the positive

x axis.
Find the components of
A
:
using Equation 1.10: A
x
5 A cos (245.08) 5 (25.0 km)(0.707) 5 17.7 km
A
y
5 A sin (245.08) 5 (25.0 km)(20.707) 5 217.7 km
The negative value of A
y
indicates that the hiker walks in the negative y direction on the  rst day. The signs of A
x
and A
y
also
are evident from Figure 1.18.
Find the components of
B
:
using Equation 1.10: B
x
5 B cos 60.08 5 (40.0 km)(0.500) 5 20.0 km
B
y
5 B sin 60.08 5 (40.0 km)(0.866) 5 34.6 km
(B) Determine the components of the hiker’s resultant displacement
R
:

for the trip. Find an expression for
R
:
in terms
of unit vectors.
SOLUTION
Use Equation 1.15 to  nd the components of the resultant R
x
5 A
x
1 B
x
5 17.7 km 1 20.0 km 5 37.7 km
displacement
R
:
5
A
:
1
B
:
:

R
y
5 A
y
1 B
y

5 217.7 km 1 34.6 km 5 17.0 km
Write the total displacement in unit-vector form:
R
:
5 (37.7
i
ˆ
1 17.0
j
ˆ
) km
Looking at the graphical representation in Figure 1.18, we estimate the position of the tower to be about (38km, 17 km),
which is consistent with the components of
R
:
in our result for the  nal position of the hiker. Also, both components
of
R
:
are positive, putting the  nal position in the  rst quadrant of the coordinate system, which is also consistent with
Figure 1.18.
What If? After reaching the tower, the hiker wishes to return to her car along a single straight line. What are the compo-
nents of the vector representing this hike? What should the direction of the hike be?
Answer The desired vector
R
:
car
is the negative of vector
R
:

:
R
:
car
5 2
R
:
5 (237.7
i
ˆ
2 17.0
j
ˆ
) km
The direction is found by calculating the angle that the vector makes with the x axis:
tan ␪ 5
R
car,y
R
car
,
x
5
217.0 km
237.7 km
5 0.450
which gives an angle of ␪ 5 204.28, or 24.28 south of west.
y (km)
x (km)
60.0°

45.0°
20 30
40
Tower
Car
0
20
10
−10
−20
E
N
S
W
Tent
A
S
B
S
R
S
Figure 1.18  (Example 1.8) The total
displacement of the hiker is the vector
R
:
5
A
:
1
B

:
.
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22CHAPTER 1
|
Introduction and Vectors
1.10
|
Modeling, Alternative Representations,
and Problem-Solving Strategy
Most courses in general physics require the student to learn the skills of problem
solving, and examinations usually include problems that test such skills. This sec-
tion describes some useful ideas that will enable you to enhance your understand-
ing of physical concepts, increase your accuracy in solving problems, eliminate
initial panic or lack of direction in approaching a problem, and organize your
work.
One of the primary problem-solving methods in physics is to form an appropriate
model of the problem. A model is a simpli ed substitute for the real problem that
allows us to solve the problem in a relatively simple way. As long as the predictions
of the model agree to our satisfaction with the actual behavior of the real system, the
model is valid. If the predictions do not agree, the model must be re ned or replaced
with another model. The power of modeling is in its ability to reduce a wide variety
of very complex problems to a limited number of classes of problems that can be
approached in similar ways.
In science, a model is very different from, for example, an architect’s scale
model of a proposed building, which appears as a smaller version of what it rep-

resents. A scienti c model is a theoretical construct and may have no visual simi-
larity to the physical problem. A simple application of modeling is presented in
Example 1.9, and we shall encounter many more examples of models as the text
progresses.
Models are needed because the actual operation of the Universe is extremely
complicated. Suppose, for example, we are asked to solve a problem about the
Earth’s motion around the Sun. The Earth is very complicated, with many processes
occurring simultaneously. These processes include weather, seismic activity, and
ocean movements as well as the multitude of processes involving human activity.
Trying to maintain knowledge and understanding of all these processes is an impos-
sible task.
The modeling approach recognizes that none of these processes affects the mo-
tion of the Earth around the Sun to a measurable degree. Therefore, these details
are all ignored. In addition, as we shall  nd in Chapter 11, the size of the Earth
does not affect the gravitational force between the Earth and the Sun; only the
masses of the Earth and Sun and the distance between them determine this force.
In a simpli ed model, the Earth is imagined to be a particle, an object with mass
but zero size. This replacement of an extended object by a particle is called the
particle model, which is used extensively in physics. By analyzing the motion of a
particle with the mass of the Earth in orbit around the Sun, we  nd that the predic-
tions of the particle’s motion are in excellent agreement with the actual motion of
the Earth.
The two primary conditions for using the particle model are as follows:
• The size of the actual object is of no consequence in the analysis of its motion.
• Any internal processes occurring in the object are of no consequence in the
analysis of its motion.
Both of these conditions are in action in modeling the Earth as a particle. Its radius
is not a factor in determining its motion, and internal processes such as thunder-
storms, earthquakes, and manufacturing processes can be ignored.
Four categories of models used in this book will help us understand and solve

physics problems. The first category is the geometric model. In this model, we
form a geometric construction that represents the real situation. We then set
aside the real problem and perform an analysis of the geometric construction.
Consider a popular problem in elementary trigonometry, as in the following
example.
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1.10
|
Modeling, Alternative Representations, and Problem-Solving Strategy 23
You may have solved a problem very similar to Example 1.9 but never thought about
the notion of modeling. From the modeling approach, however, once we draw the
triangle in Figure 1.19, the triangle is a geometric model of the real problem; it is a
substitute. Until we reach the end of the problem, we do not imagine the problem to
be about a tree but to be about a triangle. We use trigonometry to  nd the vertical leg
of the triangle, leading to a value of 23.3 m. Because this leg represents the height of the
tree, we can now return to the original problem and claim that the height of the tree
is 23.3 m.
Other examples of geometric models include modeling the Earth as a perfect
sphere, a pizza as a perfect disk, a meter stick as a long rod with no thickness, and an
electric wire as a long, straight cylinder.
The particle model is an example of the second category of models, which we will call
simpli cation models. In a simpli cation model, details that are not signi cant in deter-
mining the outcome of the problem are ignored. When we study rotation in Chapter 10,
objects will be modeled as rigid objects. All the molecules in a rigid object maintain their
exact positions with respect to one another. We adopt this simpli cation model because
a spinning rock is much easier to analyze than a spinning block of gelatin, which is not

a rigid object. Other simpli cation models will assume that quantities such as friction
forces are negligible, remain constant, or are proportional to some power of the object’s
speed.
The third category is that of analysis models, which are general types of problems
that we have solved before. An important technique in problem solving is to cast a
new problem into a form similar to one we have already solved and which can be used
as a model. As we shall see, there are about two dozen analysis models that can be
used to solve most of the problems you will encounter. We will see our  rst analysis
models in Chapter 2, where we will discuss them in more detail.
The fourth category of models is structural models. These models are generally
used to understand the behavior of a system that is far different in scale from our mac-
roscopic world—either much smaller or much larger—so that we cannot in teract
with it directly. As an example, the notion of a hydrogen atom as an electron in a
circular orbit around a proton is a structural model of the atom. We will discuss this
model and structural models in general in Chapter 11.
Example 1.9
|
Finding the Height of a Tree
You wish to  nd the height of a tree but cannot measure it directly. You stand 50.0 m from the tree and determine that a
line of sight from the ground to the top of the tree makes an angle of 25.08 with the ground. How tall is the tree?
SOLUTION
Figure 1.19 shows the tree and a right triangle corresponding
to the information in the problem superimposed over it. (We
assume that the tree is exactly perpendicular to a perfectly  at
ground.) In the triangle, we know the length of the horizontal
leg and the angle between the hypotenuse and the horizon-
tal leg. We can  nd the height of the tree by calculating the
length of the vertical leg. We do so with the tangent function:
tan ␪ 5
opposite si

d
e
adjacent side
5
h
50.0 m
h 5 (50.0 m) tan ␪ 5 (50.0 m) tan 25.08 5 23.3 m
h
25.0°
50.0 m
Figure 1.19 (Example 1.9) The height of a tree can be found
by measuring the distance from the tree and the angle of sight to
the top above the ground. This problem is a simple example of
geometrically modeling the actual problem.
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24CHAPTER 1
|
Introduction and Vectors
Intimately related to the notion of modeling is that of forming alternative repre-
sentations of the problem. A representation is a method of viewing or presenting the
information related to the problem. Scientists must be able to communicate com-
plex ideas to individuals without scienti c backgrounds. The best representation to
use in conveying the information successfully will vary from one individual to the
next. Some will be convinced by a well-drawn graph, and others will require a picture.
Physicists are often persuaded to agree with a point of view by examining an equation,
but nonphysicists may not be convinced by this mathematical representation of the

information.
A word problem, such as those at the ends of the chapters in this book, is one
representation of a problem. In the “real world” that you will enter after graduation,
the initial representation of a problem may be just an existing situation, such as the
effects of global warming or a patient in danger of dying. You may have to identify
the important data and information, and then cast the situation yourself into an
equivalent word problem!
Considering alternative representations can help you think about the information
in the problem in several different ways to help you understand and solve it. Several
types of representations can be of assistance in this endeavor:
• Mental representation. From the description of the problem, imagine a scene
that describes what is happening in the word problem, then let time progress so
that you understand the situation and can predict what changes will occur in the
situation. This step is critical in approaching every problem.
• Pictorial representation. Drawing a picture of the situation described in the
wordproblem can be of great assistance in understanding the problem. In
Example 1.9, the pictorial representation in Figure 1.19 allows us to identify the
triangle as a geometric model of the problem. In architecture, a blueprint is a
pictorial representation of a proposed building.
Generally, a pictorial representation describes what you would see if you were
observing the situation in the problem. For example, Figure 1.20 shows a
pictorial representation of a baseball player hitting a short pop foul. Any coor-
dinate axes included in your pictorial representation will be in two dimensions:
x and y axes.
• Simpli ed pictorial representation. It is often useful to redraw the picto-
rial representation without complicating details by applying a simpli cation
model. This process is similar to the discussion of the particle model described
earlier. In a pictorial representation of the Earth in orbit around the Sun, you
might draw the Earth and the Sun as spheres, with possibly some attempt to
draw continents to identify which sphere is the Earth. In the simpli ed picto-

rial representation, the Earth and the Sun would be drawn simply as dots,
representing particles. Figure 1.21 shows a simpli ed pictorial representation
corresponding to the pictorial representation of the baseball trajectory in
Figure 1.20. The notations v
x
and v
y
refer to the components of the velocity
vector for the baseball. We shall use such simpli ed pictorial representations
throughout the book.
• Graphical representation. In some problems, drawing a graph that describes the
situation can be very helpful. In mechanics, for example, position–time graphs
can be of great assistance. Similarly, in thermodynamics, pressure–volume
graphs are essential to understanding. Figure 1.22 shows a graphical represen-
tation of the position as a function of time of a block on the end of a vertical
spring as it oscillates up and down. Such a graph is helpful for understanding
simple harmonic motion, which we study in Chapter 12.
A graphical representation is different from a pictorial representation, which
is also a two-dimensional display of information but whose axes, if any, represent
length coordinates. In a graphical representation, the axes may represent any
two related variables. For example, a graphical representation may have axes for
temperature and time. Therefore, in comparison to a pictorial representation, a
Figure 1.20 A pictorial
representation of a pop foul being hit
by a baseball player.
v
y
v
x
v

S
Figure 1.21 A simpli ed pictorial
representation for the situation
shown in Figure 1.20.
y
t
Figure 1.22 A graphical
representation of the position as a
function of time of a block hanging
from a spring and oscillating.
10293_ch01_001-034.indd 2410293_ch01_001-034.indd 24 1/6/12 2:35 PM1/6/12 2:35 PM
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deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Licensed to:
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1.10
|
Modeling, Alternative Representations, and Problem-Solving Strategy 25
GENERAL PROBLEM-SOLVING STRATEGY
Conceptualize

• The  rst things to do when approaching a problem are to think about and
understand the situation. Study carefully any representations of the informa-
tion (for example, diagrams, graphs, tables, or photographs) that accompany
the problem. Imagine a movie, running in your mind, of what happens in the
problem.
• If a pictorial representation is not provided, you should almost always make a
quick drawing of the situation. Indicate any known values, perhaps in a table
or directly on your sketch.
• Now focus on what algebraic or numerical information is given in the prob-

lem. Carefully read the problem statement, looking for key phrases such as
“starts from rest” (v
i
5 0) or “stops” (v
f
5 0).
• Now focus on the expected result of solving the problem. Exactly what is
the question asking? Will the  nal result be numerical or algebraic? Do you
know what units to expect?
• Don’t forget to incorporate information from your own experiences and com-
mon sense. What should a reasonable answer look like? For example, you
wouldn’t expect to calculate the speed of an automobile to be 5 3 10
6
m/s.
Categorize
• Once you have a good idea of what the problem is about, you need to simplify
the problem. Remove the details that are not important to the solution. For
example, model a moving object as a particle. If appropriate, ignore air resis-
tance or friction between a sliding object and a surface.
• Once the problem is simpli ed, it is important to categorize the problem. Is it a
simple substitution problem such that numbers can be substituted into an equa-
tion? If so, the problem is likely to be  nished when this substitution is done.
If not, you face what we call an analysis problem: the situation must be analyzed
more deeply to reach a solution.
• If it is an analysis problem, it needs to be categorized further. Have you
seen this type of problem before? Does it fall into the growing list of types
of problems that you have solved previously? If so, identify any analysis
graphical representation is generally not something you would see when observ-
ing the situation in the problem with your eyes.
• Tabular representation. It is sometimes helpful to organize the information in

tabular form to help make it clearer. For example, some students  nd that mak-
ing tables of known quantities and unknown quantities is helpful. The periodic
table of the elements is an extremely useful tabular representation of informa-
tion in chemistry and physics.
• Mathematical representation. The ultimate goal in solving a problem is often the
mathematical representation. You want to move from the information contained
in the word problem, through various representations of the problem that allow
you to understand what is happening, to one or more equations that represent
the situation in the problem and that can be solved mathematically for the desired
result.
Besides what you might expect to learn about physics concepts, a very valuable skill
you should acquire from your physics course is the ability to solve complicated prob-
lems. The way physicists approach complex situations and break them into manage-
able pieces is extremely useful. The following is a general problem-solving strategy to
guide you through the steps. To help you remember the steps of the strategy, they are
Conceptualize, Categorize, Analyze, and Finalize.
10293_ch01_001-034.indd 2510293_ch01_001-034.indd 25 1/6/12 2:35 PM1/6/12 2:35 PM
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deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Licensed to:
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