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NAME ______________________________________ DATE _______________ CLASS ____________________
Two-Dimensional Motion and Vectors
Problem A
FINDING RESULTANT MAGNITUDE AND DIRECTION
PROBLEM
Cheetahs are, for short distances, the fastest land animals. In the course
of a chase, cheetahs can also change direction very quickly. Suppose a
cheetah runs straight north for 5.0 s, quickly turns, and runs 3.00 × 102 m
west. If the magnitude of the cheetah’s resultant displacement is 3.35 ×
102 m, what is the cheetah’s displacement and velocity during the first
part of its run?
SOLUTION
1. DEFINE
Given:
∆t1 = 5.0 s
∆x = 3.00 s × 102 m
d = 3.35 × 102 m
Unknown:
∆y = ?
vy = ?
∆ x = 3.00 × 102 m
Diagram:
∆y = ?
N
d = 3.35 × 102 m
Copyright © by Holt, Rinehart and Winston. Allrights reserved.
2. PLAN
Choose the equation(s) or situation: Use the Pythagorean theorem to subtract
one of the displacements at right angles from the total displacement, and thus
determine the unknown component of displacement.
d2 = ∆x2 + ∆y2
Use the equation relating displacement to constant velocity and time, and use the
calculated value for ∆y and the given value for ∆t to solve for v.
∆y
∆t
∆v =
Rearrange the equation(s) to isolate the unknown(s):
∆y 2 = d 2 – ∆x2
∆y =
vy =
3. CALCULATE
d 2
–
∆
x2
∆y
∆t
Substitute the values into the equation(s) and solve: Because the value for ∆y
is a displacement magnitude, only the positive root is used (∆y > 0).
∆y = (3
02
m
02
m
.3
5×1
)2−(3.
00
×1
)2
= 1.
05
m2−
0×
04
m2
12
×1
9.0
1
Problem A
17
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NAME ______________________________________ DATE _______________ CLASS ____________________
= 2.
2×
04
m
1
= 1.5 × 102 m, north
1.5 × 102 m
vy =  = 3.0 × 101 m/s, north
5.0 s
4. EVALUATE
The cheetah has a top speed of 30 m/s, or 107 km/h. This is equal to about
67 miles/h.
ADDITIONAL PRACTICE
1. An ostrich cannot fly, but it is able to run fast. Suppose an ostrich runs
east for 7.95 s and then runs 161 m south, so that the magnitude of the
ostrich’s resultant displacement is 226 m. Calculate the magnitude of the
ostrich’s eastward component and its running speed.
2. The pronghorn antelope, found in North America, is the best longdistance runner among mammals. It has been observed to travel at an average speed of more than 55 km/h over a distance of 6.0 km. Suppose the
antelope runs a distance of 5.0 km in a direction 11.5° north of east,
turns, and then runs 1.0 km south. Calculate the resultant displacement.
4. In 1926, Gertrude Ederle of the United States became the first woman to
swim across the English channel. Suppose Ederle swam 25.2 km east
from the coast near Dover, England, then made a 90° turn and traveled
south for 21.3 km to a point east of Calais, France. What was Ederle’s resultant displacement?
5. The emperor penguin is the best diver among birds: the record dive is
483 m. Suppose an emperor penguin dives vertically to a depth of 483 m
and then swims horizontally a distance of 225 m. What angle would the
vector of the resultant displacement make with the water’s surface? What
is the magnitude of the penguin’s resultant displacement?
6. A killer whale can swim as fast as 15 m/s. Suppose a killer whale swims in
one direction at this speed for 8.0 s, makes a 90° turn, and continues
swimming in the new direction with the same speed as before. After a certain time interval, the magnitude of the resultant displacement is 180.0 m.
Calculate the amount of time the whale swims after changing direction.
7. Woodcocks are the slowest birds: their average speed during courtship
displays can be as low as 8.00 km/h. Suppose a woodcock flies east for
15.0 min. It then turns and flies north for 22.0 min. Calculate the magnitude of the resultant displacement and the angle between the resultant
displacement and the woodcock’s initial displacement.
18
Holt Physics Problem Workbook
Copyright © by Holt, Rinehart and Winston. Allrights reserved.
3. Kangaroos can easily jump as far 8.0 m. If a kangaroo makes five such
jumps westward, how many jumps must it make northward to have a
northwest displacement with a magnitude of 68 m? What is the angle of
the resultant displacement with respect to north?
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Problem Workbook Solutions
Two-Dimensional
Motion and Vectors
Additional Practice A
Givens
1. ∆tx = 7.95 s
Solutions
d 2 = ∆x 2 + ∆y 2
∆y = 161 m
∆x = d 2−
∆y2 = (2
m
)2
−(16
)2 = 5.
04
m2−
04
m2
26
1m
11
×1
2.5
9×1
d = 226 m
∆x = 2.
04
m2 = 159 m
52
×1
∆x = 159 m
∆x 159 m
v =  =  = 20.0 m/s
∆tx 7.95 s
2. d1 = 5.0 km
θ1 = 11.5°
2
d = 1.0 km
q2 = −90.0°
∆xtot = d1(cos q1) + d2(cos q2) = (5.0 km)(cos 11.5°) + (1.0 km)[cos(−90.0°)]
II
∆xtot = 4.9 km
∆ytot = d1(sin q1) + d2(sin q2) = (5.0 km)(sin 11.5°) + (1.0 km)[sin(−90.0°)]
= 1.0 km − 1.0 km
∆ytot = 0.0 km
d = (∆
xtot
)2
+(∆
ytot
)2 = (4
)2
+(0.
)2
.9
km
0km
d = 4.9 km
∆ytot
0.0 km
= tan−1  = 0.0°, or due east
q = tan−1 
∆xtot
4.9 km
Copyright © by Holt, Rinehart and Winston. All rights reserved.
3. ∆x = 5 jumps
1 jump = 8.0 m
d = 68 m
d 2 = ∆x 2 + ∆y 2
∆y = d 2−
∆x2 = (6
)2
−[(5
03
m2−
03
m2
8m
)(
8.
0m
)]
2 = 4.
6×1
1.6
×1
∆y = 3.
m = 55 m
0×103
55 m
number of jumps northward =  = 6.9 jumps = 7 jumps
8.0 m/jump
∆x
(5)(8.0 m)
q = tan−1  = tan−1  = 36° west of north
∆y
55 m
4. ∆x = 25.2 km
∆y = 21.3 km
d= ∆
∆y2 = (2
)2
+(21
)2
x2+
5.
2km
.3
km
d = 63
5km
2+454
km
2 = 10
89
km
2
d = 33.00 km
∆y
21.3 km
q = tan−1  = tan−1 
∆x
25.2 km
q = 42.6° south of east
Section Two — Problem Workbook Solutions
II Ch. 3–1
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Givens
Solutions
5. ∆y = −483 m
∆x = 225 m
∆y
−483
q = tan−1  = tan−1  = −65.0° = 65.0° below the waters surface
∆x
225 m
2
d= ∆
∆y2 = (2
m
)2
+(−
)2
x+
25
48
3m
d = 5.
m2+
m2 = 2.
m2
06
×104
2.3
3×105
84
×105
d = 533 m
6. v = 15.0 m/s
∆tx = 8.0 s
d = 180.0 m
d 2 = ∆x 2 + ∆y 2 = (v∆tx)2 + (v∆ty)2
d 2 = v 2(∆tx2 + ∆ty2)
∆ty =
180.0 m 2
 − (8.0 s)2 = 14
s2
−64
s2 = 8.
01s2
4
0×1
15.0 m/s
d

v
2
− ∆tx2 =
∆ty = 8.9 s
7. v = 8.00 km/h
∆tx = 15.0 min
II
∆ty = 22.0 min
d= ∆
∆y2 = (v
tx
)2
+(v∆
ty
)2
x2+
∆
=v ∆
tx2+
∆
ty2
(15.0min) +(22.0min)
1h d = (8.00 km/h)  22
60 min 5min+484min
8.00 km d =  70
60 min 9min = 3.55 km
v∆t
∆t
∆y
22.0 min
q = tan  = tan  = tan  = tan 
∆x v∆t ∆t 15.0 min
1h
d = (8.00 km/h) 
60 min
2
2
2
2
2
−1
−1
−1
y
x
−1
y
x
q = 55.7° north of east
1. d = (5)(33.0 cm)
∆y = 88.0 cm
88.0 cm
∆y
q = sin−1  = sin−1  = 32.2° north of west
(5)(33.0 cm)
d
∆x = d(cos q ) = (5)(33.0 cm)(cos 32.2°) = 1.40 × 102 cm to the west
2. q = 60.0°
d = 10.0 m
3. d = 10.3 m
∆y = −6.10 m
∆x = d(cos q) = (10.0 m)(cos 60.0°) = 5.00 m
∆y = d(sin q ) = (10.0 m)(sin 60.0°) = 8.66 m
Finding the angle between d and the x-axis yields,
∆y
−6.10 m
q1 = sin−1  = sin−1  = −36.3°
d
10.3 m
The angle between d and the negative y-axis is therefore,
q = −90.0 − (−36.3°) = −53.7°
q = 53.7° on either side of the negative y-axis
d 2 + ∆x 2 + ∆y2
∆x = d 2−
∆
y2 = (1
)2
−(−
m)2 = 10
m2
0.
3m
6.
10
6m
2−37.
2m
2 = 69
∆x = ±8.3 m
II Ch. 3–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Additional Practice B