HOMEWORK CHAPTER 1
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 the quantity
1 2
at 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 the quantity
1 3
at 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
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 )
Fs I
m G J
Hm K
2
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.
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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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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REASONING We use the facts that 1 mi = 5280 ft, 1 m = 3.281 ft, and
1 yd = 3 ft. With these facts we construct three conversion factors: (5280 ft)/(1 mi)
= 1, (1 m)/(3.281 ft) = 1, and (3 ft)/(1 yd) = 1.
SOLUTION By multiplying by the given distance d of the fall by the appropriate
conversion factors we find that
F
5280 ft IF 1 m I
3 ft IF 1 m I
b
551 yd g
b gF
G
J
G
J
G
H1 mi KH3.281 ft K
H3.281 ft J
K
H1 yd J
KG
d  6 mi
10 159 m
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112. REASONING AND SOLUTION The
angle  is given by tan  = ho/ha so

 = tan–1(12.0 m/100.0 m) =
ho

6.84
h
a
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17. REASONING AND SOLUTION Consider the following views of the cube.
Bottom
View
Side View
Na
L
c
a
Na
Cl
a
a
Cl
Na
L
Na
The length, L, of the diagonal of the bottom face of the cube can be found using the
Pythagorean theorem to be
2
2
2
2
L = a + a = 2(0.281 nm) = 0.158 nm
2
or
L = 0.397 nm
The required distance c is also found using the Pythagorean theorem.
c2 = L2 + a2 = (0.397 nm)2 + (0.281 nm)2 = 0.237 nm2
Then,
c  0.487 nm
18.
REASONING AND SOLUTION
 = tan-1(a/L) = tan-1(0.281/0.397) =
35.3
28. REASONING The displacement vector R for a straight-line
dash between the starting and finishing points can be found by
drawing the vectors A, B, and C to scale in both magnitude and Side of court
orientation, in a tail-to-head fashion, and connecting the tail of A
to the head of C. The magnitude of the resultant can be found
by measuring its length and making use of the scale factor.
Similarly, the direction of the resultant is found by measuring
the angle  it makes with the side of the court.
SOLUTION By construction and measurement, the magnitude
of the resultant R is 20 m . The angle  made with the side of
C
R
B

A
the court is 15° .
35.
SSM REASONING The x and y components of r are mutually perpendicular;
therefore, the magnitude of r can be found using the Pythagorean theorem. The
direction of r can be found using the definition of the tangent function.
SOLUTION According to the Pythagorean theorem, we have
r
x 2  y 2  ( 125 m) 2  ( 184 m) 2  222 m
The angle  is
184 m 
 55.8
125 m 
  tan 1 
40. REASONING AND SOLUTION The first three rows of the following table gives
the components of each of the three individual displacements. The fourth row gives
the components of the resultant displacement. The directions due east and due north
have been taken as the positive directions.
Displacement
East/West
Component
North/South
Component
A
B
C
–52 paces
–(42 paces) cos 30.0° = –36 paces
0
0
(42 paces) sin 30.0° = 21 paces
25 paces
R=A+B+C
–88 paces
46 paces
a. Therefore, the magnitude of the displacement in the direction due north is
46 paces .
b.
Similarly, the magnitude of the displacement in the direction due west is
88 paces .
46. REASONING AND SOLUTION Since the resultant of the three vectors is zero, each
component of the resultant must be equal to zero.
The x components of the first two forces are
F1x = (166 N) cos 60.0° = 83.0 N
F2x = (284 N) cos 30.0° = 246 N
The x component of the resultant of the three vectors is, therefore,
(83 N) + (246 N) + F3x = 0
or F3x = – 329 N. The minus sign indicates that F3x points in the negative x
direction.
The y components of the first two forces are
F1y = (166 N) sin 60.0° = 144 N
F2y = (284 N) sin 30.0° = 142 N
The y component of the resultant of the three vectors is, therefore,
(144 N) + (142 N) + F3y = 0
or F3y = – 286 N. The minus sign indicates that F3y points in the negative y
direction.
The magnitude of F3 is, from
the Pythagorean theorem,
y
( 329 N) 2 + ( 286 N) 2 = 436 N
The angle made by F3 with the
negative x axis is
  tan
1
286 N O
L
M
N392 N P
Q 41.0
F
3x
F
3y

x
F
3
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