CHAPTER 1 INTRODUCTION AND
MATHEMATICAL CONCEPTS
CONCEPTUAL QUESTIONS
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1.
REASONING AND SOLUTION
a. The SI unit for x is m. The SI units for the quantity vt are
F
IJ
G
HK
m
m

 
(s)  m
(s)  m
 s 
s
Therefore, the units on the left hand side of the equation are consistent with the units on the
right hand side.
b. As described in part a, the SI units for the quantities x and vt are both m. The SI units for
1
the quantity at2 are
2
mI
F
G
Hs J
K(s )  m
2
2
Therefore, the units on the left hand side of the equation are consistent with the units on the
right hand side.
c. The SI unit for v is m/s. The SI unit for the quantity at is
mI
m
F
(s) 
G
J
Hs K s
2
Therefore, the units on the left hand side of the equation are consistent with the units on the
right hand side.
d. As described in part c, the SI units of the quantities v and at are both m/s. The SI unit of
1
the quantity at3 is
2
mI
F
G
Hs J
K(s )  m  s
3
2
Thus, the units on the left hand side are not consistent with the units on the right hand side.
In fact, the right hand side is not a valid operation because it is not possible to add physical
quantities that have different units.
e. The SI unit for the quantity v3 is m3/s3. The SI unit for the quantity 2ax3 is
2
INTRODUCTION AND MATHEMATICAL CONCEPTS
mI
F
G
Hs J
K(m
2
2
)
m3
s2
Therefore, the units on the left hand side of the equation are not consistent with the units on
the right hand side.
f. The SI unit for the quantity t is s. The SI unit for the quantity
m

(m / s 2 )
F
s I
G
Hm J
K
2
m
2x
is
a
s2  s
Therefore, the units on the left hand side of the equation are consistent with the units on the
right hand side.
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2.
REASONING AND SOLUTION
The units of the ratio x/v are
The quantity tan  is dimensionless and has no units.
F
I
G
HJ
K
m
s
m
s
(m / s)
m
Thus, the units on the left side of the equation are not consistent with those on the right side,
and the equation tan  = x/v is not a possible relationship between the variables x, v, and .
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3.
REASONING AND SOLUTION It is not always possible to add two numbers that have
the same dimensions. In order to add any two physical quantities they must be expressed in
the same units. Consider the two lengths: 1.00 m and 1.00 cm. Both quantities are lengths
and, therefore, have the dimension [L]. Since the units are different, however, these two
numbers cannot be added.
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4.
REASONING AND SOLUTION
a. The dimension of a physical quantity describes the physical nature of the quantity and the
kind of unit that is used to express the quantity. It is possible for two quantities to have the
same dimensions but different units. All lengths, for example, have the dimension [L].
However, a length may be expressed in any length unit, such as kilometers, meters,
centimeters, millimeters, inches, feet or yards.
As another illustration, the quantities 100 g and 1.5 kg are masses and have the
dimensions [M]; however, they have different units.
b. All quantities with the same units must have the same dimensions. For example, all
quantities expressed in kilograms have the dimension [M]; all quantities expressed in
meters have the dimensions [L].
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Chapter 1 Conceptual Questions
5.
3
REASONING AND SOLUTION For the equation to be valid, the dimensions of the left
hand side of the equation must be the same as the dimensions on the right hand side. Since
the quantity c has no dimensions, it does not contribute to the dimensions of the right hand
side, regardless of the value of n. Therefore, the value of n cannot be determined from
dimensional analysis.
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6.
REASONING AND SOLUTION The following table shows the value of sin , cos , the
ratio (sin )/(cos ) and tan .

30.0
39.0
53.0
60.0
sin 
0.500
0.629
0.799
0.866
cos 
(sin )/(cos )
0.866
0.777
0.602
0.500
0.577
0.810
1.33
1.73
tan 
0.577
0.810
1.33
1.73
From the definitions given in Equations 1.1-1.3, we have
sin  ho / h ho


 tan 
cos  ha / h ha
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7.
REASONING AND SOLUTION
a. The graph below shows sin  plotted on the vertical axis and  on the horizontal axis,
with  in 15° increments from  = 0 to  = 720°.
4
INTRODUCTION AND MATHEMATICAL CONCEPTS
b.
The graph below shows cos  plotted on the vertical axis and  on the horizontal axis, with 
in 15° increments from  = 0 to  = 720°.
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8.
REASONING AND SOLUTION A vector quantity has both magnitude and direction. The
number of people attending a football game, the number of days in a month, and the number
of pages in a book can all be completely specified by giving a magnitude only. Hence, none
of these quantities can be considered a vector.
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9.
REASONING AND SOLUTION For two vectors to be equal, they must be equal in
magnitude and have the same direction. Only vectors A, B, and D have the same magnitude.
These three vectors are shown below:
North
North
North
A
30°
60°
East
East
D
East
30°
B
Clearly, only vectors A and D have the same magnitude and point in the same direction;
hence, A and D are equal.
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Chapter 1 Conceptual Questions
5
10. REASONING AND SOLUTION For two vectors to be equal, they must be equal in
magnitude and have the same direction. Thus, two vectors with the same magnitude are not
necessarily equal. They must also point in the same direction.
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11. REASONING AND SOLUTION One can arrive back at the starting point after making
eight consecutive displacements that add to zero only if one traverses three of the edges on
one face and three edges on the opposite face (six displacements; the remaining two
displacements occur in going from one opposite face to the other). For any particular
starting point, there are four independent ways to traverse three edges on opposite faces.
These are illustrated in the figures below. In (A) and (B), one traverses three consecutive
edges before moving to the opposite face; in (C) and (D), one traverses two consecutive
edges, moves to the opposite face, traverses three consecutive edges, then moves back to the
original face and traverses the third edge on that face ending at the starting point.
(A)
starting point
(C)
starting point
(B)
(D)
starting point
starting point
For any particular starting point, one can initially move in any one of three possible
directions. Thus, there are a total of 4 (independent ways)  3 (possible directions) or 12
ways to arrive back at the starting point that involve eight displacement vectors.
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12. REASONING AND SOLUTION
(a) Any vector has a zero component in the direction that is perpendicular to the direction
of that vector. Thus, it is possible for one component of a vector to be zero, while the vector
itself is not zero.
6
INTRODUCTION AND MATHEMATICAL CONCEPTS
(b) In order for a vector to be zero, its components in all mutually perpendicular directions
must also be zero. Therefore, it is not possible for a vector to be zero, while one of its
components is not zero.
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13. REASONING AND SOLUTION If two or more vectors have a resultant of zero, then,
when they are arranged in a tail-to-head fashion the head of the last vector must touch the
tail of the first vector. In order to satisfy this requirement with two vectors, the vectors
must be equal in magnitude and opposite in direction. Therefore, it is not possible to add
two perpendicular vectors so that the vector sum is zero.
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14. REASONING AND SOLUTION Three or more vectors with
unequal magnitudes may add together so that their vector sum is
zero. To do so, they must be oriented so that, when they are
placed in a tail-to-head fashion, the head of the last vector
touches the tail of the first vector, as in the drawing at the right.
C
B
A
A+B+C= 0
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15. REASONING AND SOLUTION The vector equation A + B = 0, implies that A = –B.
a. The magnitude of A must be equal to the magnitude of B.
b. The vectors A and B must point in opposite directions, as indicated by the minus sign in
A = –B.
The same conclusions can be reached from a geometric argument. If A + B = 0, then, when
A and B are placed tail-to-head, the head of B must touch the tail of A. Thus, the vectors A
and B must have the same length (i.e., the same magnitude) and point in opposite directions.
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16. REASONING AND SOLUTION The equation A + B = C tells us that the vector C is the
resultant of the vectors A and B. The magnitudes of the vectors A, B, and C are related by
A2 + B 2 = C 2. This has the same form as the Pythagorean theorem that relates the length of
the two sides of a right triangle and the length of the hypotenuse. Thus, the vectors A and B
must be at right angles (or perpendicular) to each other.
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17. REASONING AND SOLUTION The equation A + B = C tells us that the vector C is the
resultant of the vectors A and B. The magnitudes of the vectors A, B, and C are related by A
+ B = C. In other words, the length of the vector C is equal to the combined lengths of
vectors A and B. Therefore, the vectors A and B must point in the same direction.
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18. REASONING AND SOLUTION If the magnitude of a vector is doubled, we may conclude
that the magnitude of each component of the vector has also doubled.
Chapter 1 Conceptual Questions
This is best explained using the arbitrary vector, A, shown at
the right. The angle , that specifies the direction of A, is
given by
  tan 1
FA I
G
HA J
K
y
y
7
A

Ax
Ay
x
x
Suppose that the magnitude of A is doubled. The vector A is now twice as long, but its
direction (and therefore the angle ) remains unchanged. In order for the angle  to remain
the same after the magnitude of A is doubled, both Ax and Ay must double, so that the ratio
Ay Ax remains the same.
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19. REASONING AND SOLUTION
a. As the angle increases from 0° to 90°, the x component of the vector decreases in
magnitude while the y component increases in magnitude. When the vector has rotated
through an angle of 90°, the x component has zero magnitude and the magnitude of the y
component is equal to the magnitude of the original vector (see figure below).
b. As the angle increases from 90° to 180°, the y component of the vector decreases in
magnitude while the x component increases in magnitude. When the vector has rotated
through an angle of 180°, the y component has zero magnitude and the magnitude of the x
component is equal to the magnitude of the original vector.
c. As the angle increases from 180° to 270°, the x component of the vector decreases in
magnitude while the y component increases in magnitude. When the vector has rotated
through an angle of 270°, the x component has zero magnitude and the magnitude of the y
component is equal to the magnitude of the original vector.
d. As the angle increases from 270° to 360°, the y component of the vector decreases in
magnitude while the x component increases in magnitude. When the vector has rotated
through an angle of 360°, the y component has zero magnitude and the magnitude of the x
component is equal to the magnitude of the original vector.
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8
INTRODUCTION AND MATHEMATICAL CONCEPTS
20. REASONING AND SOLUTION In general, a vector that has a component of zero along
the x axis of a certain axes system does not have a component of zero along the x axis of
another (rotated) axes system. Consider the vector A shown below.
y
y A
x'
y'
A


x
Figure 1
A x'
x
Figure 2
Relative to the x, y axes shown in Figure 1, the vector A has a component of zero along the
x axis. Figure 2 shows the same vector along with a new axes system x', y'. The primed axes
system is rotated at an arbitrary angle  counterclockwise relative to the unprimed x, y axes
system. Clearly the component of A along the x' axis is non zero for most arbitrary values of
the (rotation) angle . The exceptions occur when  = 0, 180° and 360°.
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