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NAME ______________________________________ DATE _______________ CLASS ____________________
Two-Dimensional Motion and Vectors
Problem C
ADDING VECTORS ALGEBRAICALLY
PROBLEM
The southernmost point in the United States is called South Point, and is
located at the southern tip of the large island of Hawaii. A plane designed
to take off and land in water leaves South Point and flies to Honolulu, on
the island of Oahu, in three separate stages. The plane first flies 83.0 km at
22.0° west of north from South Point to Kailua Kona, Hawaii. The plane
then flies 146 km at 21.0° west of north from Kailua Kona to Kahului, on
the island of Maui. Finally, the plane flies 152 km at 17.5° north of west
from Kahului to Honolulu. What is the plane’s resultant displacement?
SOLUTION
1. DEFINE
Given:
d1 = 83.0 km
d2 = 146 km
d3 = 152 km
q 1 = 22.0° west of north
q 2 = 21.0° west of north
q 3 = 17.5° north of west
Unknown:
d=?
q =?
Diagram:
N
d 3 = 152 km
θ 3 = 17.5°
d
θ 2 = 21.0°
d1 = 83.0 km
θ1 = 22.0°
θ
2. PLAN
Choose the equation(s) or situation: Express the components of each vector in
terms of sine or cosine functions.
∆x1 = d1 (sin q 1)
∆y1 = d1 (cos q 1)
∆x2 = d2 (sin q 2)
∆y2 = d2 (cos q 2)
∆x3 = d3 (cos q 3)
∆y3 = d3 (sin q 3)
Note that the angles q 1 and q 2 are with respect to the y axis (north), and so the x
components are in terms of sin q. Write the equations for ∆xtot and ∆ytot , the
components of the total displacement.
Ch. 3–6
Holt Physics Problem Bank
Copyright © by Holt, Rinehart and Winston. All rights reserved.
d 2 = 146 km
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NAME ______________________________________ DATE _______________ CLASS ____________________
∆xtot = ∆x1 + ∆x2 + ∆x3
= d1 (sin q 1) + d 2(sin q 2) + d 3 (cos q 3)
∆ytot = ∆y1 + ∆y2 + ∆y3
= d1 (cos q 1) + d 2(cos q 2) + d 3 (sin q 3)
Use the components of the total displacement, the Pythagorean theorem, and the
tangent function to calculate the total displacement.
d = (∆
xtot
)2苶
+苶(∆
ytot
)2
苶苶
苶苶
苶苶
苶苶
冢 冣
∆ytot
q = tan−1 ᎏ
∆xtot
3. CALCULATE
Substitute the values into the equation(s) and solve:
∆xtot = (83.0 km)(sin 22.0°) + (146 km)(sin 21.0°) + (152 km)
(cos 17.5°)
= 31.1 km + 52.3 km + 145 km
= 228 km
∆ytot = (83.0 km)(cos 22.0°) + (146 km)(cos 21.0°) + (152 km)
(sin 17.5°)
= 259 km
d = 苶
(2苶
28苶km
)2苶
+苶(25
9 km
)2 =
苶苶
苶苶
苶苶
4
2
5.苶
20苶
×苶1苶
0 苶km
04苶km
04苶km
苶
苶苶+苶苶6.7
苶1苶×苶苶1苶
苶2苶 = 11
苶.9
苶1苶×苶苶1苶
苶2苶
= 345.1 km
冢
冣
259 km
q = tan−1 ᎏ = 48.6° north of west
228 km
Copyright © by Holt, Rinehart and Winston. All rights reserved.
4. EVALUATE
If the diagram is drawn to scale, compare the calculated results to the drawing.
The length of the drawn resultant is fairly close to the scaled magnitude for d,
while the angle appears to be slightly greater than 45°.
ADDITIONAL PRACTICE
1. U.S. Highway 212 extends 55 km at 37° north of east between Newell and
Mud Butte, South Dakota. It then continues for 66 km nearly due east
from Mud Butte to Faith, South Dakota. If you drive along this part of
U.S. Highway 212, what will be your total displacement?
2. Wrigley Field is one of only three original major-league baseball fields
that are still in use today. Suppose you want to drive to Wrigley Field
from the corner of 55th Street and Woodlawn Avenue, about 14 miles
south of Wrigley Field. Although not the fastest or most direct route, the
most straightforward way to reach Wrigley Field is to drive 4.1 km west
on 55th Street to Halsted Street, then turn north and drive 17.3 km on
Halsted until you reach Clark Street. Turning on Clark, you will reach
Wrigley Field after traveling 1.2 km at an angle of 24.6° west of north.
What is your resultant displacement?
Problem C
Ch. 3–7
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3. A bullet traveling 850 m ricochets from a rock. The bullet travels another 640 m, but at an angle of 36° from its previous forward motion.
What is the resultant displacement of the bullet?
4. The cable car system in San Francisco is the last of its kind that is still
in use in the United States. It was originally designed to transport large
numbers of people up the steep hills on which parts of the city are
built. If you ride seven blocks on the Powell Street cable car from the
terminal at Market Street to Pine Street, you will travel 2.00 × 102 m on
level ground, then 3.00 × 102 m at an incline of 3.0° to the horizontal,
and finally 2.00 × 102 m at 8.8° to the horizontal. What will be your resultant displacement?
5. An Arctic tern flying to Antarctica encounters a storm. The tern
changes direction to fly around the storm. If the tern flies 46 km at 15°
south of east, 22 km at 13° east of south, and finally 14 km at 14° west
of south, what is the tern’s resultant displacement?
7. The city of Amsterdam, in the Netherlands, has several canals that connect different sections of the city. Suppose you take a barge trip to the
harbor, starting at a point near the northwest corner of the Vondelpark. You would sail 2.50 × 103 m at 58.5° north of east, 375 m at 21.8°
north of east, and 875 m at 21.5° east of north. What would be your resultant displacement?
8. The elevated train, or “L,” in Chicago is a major source for mass transit
in that city. One of the lines extends from Jefferson Park, in the northwest part of town, to the Clark Street station downtown. The route of
this line runs 5.0 km at 36.9° south of east, 1.5 km due south, 8.5 km at
42.2° south of east, and 0.8 km due east. What is the resultant displacement of an “L” train from Jefferson Park to Clark Street?
9. A billiard table is positioned with its long side parallel to north. A cue
ball is then shot so that it travels 1.41 m at an angle of 45.0° west of
north, is deflected by the table’s left side, and continues to move 1.98 m
east of north at an angle of 45.0°. The ball is then deflected by the
table’s right side, so that it moves 0.42 m west of north at an angle of
45.0°. After a reflection on the north end of the table, the ball travels
1.56 m at an angle of 45.0° south of west. Determine the resultant displacement of the cue ball.
Ch. 3–8
Holt Physics Problem Bank
Copyright © by Holt, Rinehart and Winston. All rights reserved.
6. A technique used to change the direction of space probes, as well as to
give them additional speed, is to use the gravitational pull of nearby
planet. This technique was first used with the Voyager probes. Voyager 2
had traveled about 6.3 × 108 km when it reached Jupiter. Jupiter’s gravity changed Voyager’s direction by 68°. The probe then traveled about
9.4 × 108 km when it reached Saturn, and its direction was changed by
94°. Voyager 2 was now redirected; it encountered Uranus after traveling
3.4 × 109 km from Saturn. Use this information to calculate the resultant displacement of Voyager 2 as it traveled from Earth to Uranus.
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NAME ______________________________________ DATE _______________ CLASS ____________________
Copyright © by Holt, Rinehart and Winston. All rights reserved.
10. Hurricane Iniki was the most destructive cyclone to have crossed the
Hawaiian Islands in the twentieth century. It’s path was also unusual: it
moved south of the islands for 790 km at an angle of 18° north of west,
then moved due west for 150 km, turned north and continued for
470 km, and finally turned back 15° east of north and moved 240 km
to cross the island of Kauai. What was the resultant displacement of
Hurricane Iniki?
Problem C
Ch. 3–9
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Givens
Solutions
8. v = 165.2 km/s
q = 32.7°
vforward = v(cos q) = (165.2 km/s)(cos 32.7°)
vforward = 139 km/s, forward
vside = v(sin q) = (165.2 km/s)(sin 32.7°)
vside = 89.2 km/s to the side
9. v = 55.0 km/h
q = 13.0° above horizontal
vy = v(sin q) = (55.0 km/h)(sin 13.0°)
vy = 12.4 km/h, upward
vx = v(cos q) = (55.0 km/h)(cos 13.0°)
vx = 53.6 km/h, forward
10. v = 13.9 m/s
vz = v(sin qv) = (13.9 m/s)(sin 26.0°)
qh = 24.0° east of north
vz = 6.09 m/s, upward
qv = 26.0° above the
horizontal
horizontal velocity = vh = v(cos qv)
vy = vh(cos qh) = v(cos qv)(cos qh) = (13.9 m/s)(cos 26.0°)(cos 24.0°)
vy = 11.4 m/s, north
vx = vh(sin qh) = v(cos qv)(sin qh) = (13.9 m/s)(cos 26.0°)(sin 24.0°)
vx = 5.08 m/s, east
Additional Practice C
1. d1 = 55 km
∆x1 = d1(cos q1) = (55 km)(cos 37°) = 44 km
q1 = 37 north of east
∆y1 = d1(sin q1) = (55 km)(sin 37°) = 33 km
d2 = 66 km
∆x2 = d2(cos q2) = (66 km)(cos 0.0°) = 66 km
q2 = 0.0° (due east)
∆y2 = d2(sin q2) = (66 km)(sin 0.0°) = 0 km
∆ytot = ∆y1 + ∆y2 = 33 km + 0 km = 33 km
d = (∆
xtot
)2苶
+苶(∆
ytot
)2 = (1
)2苶
+苶(33
)2
苶苶
苶苶
苶苶
苶苶
苶10
苶苶km
苶苶
苶苶km
苶苶
= 1.
04苶km
03苶km
04苶km
苶21
苶苶×苶1苶
苶2苶+苶苶1.1
苶苶×苶1苶
苶2= 1.
苶32
苶苶×苶1苶
苶2苶
d = 115 km
冢 冣
冢
冣
∆ytot
33 km
q = tan−1 
= tan−1 
∆xtot
110 km
q = 17° north of east
V
V Ch. 3–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆xtot = ∆x1 + ∆x2 = 44 km + 66 km = 110 km
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Givens
2. d1 = 4.1 km
Solutions
∆x1 = d1(cos ∆1) = (4.1 km)(cos 180°) = −4.1 km
q1 = 180° (due west)
∆y1 = d1(sin q1) = (4.1 km)(sin 180°) = 0 km
d2 = 17.3 km
∆x2 = d2(cos q2) = (17.3 km)(cos 90.0°) = 0 km
q2 = 90.0° (due north)
∆y2 = d2(sin q2) = (17.3 km)(sin 90.0°) = 17.3 km
d3 = 1.2 km
q3 = 24.6° west of north
= 90.0° + 24.6° = 114.6°
∆x3 = d3(cos q3) = (1.2 km)(cos 114.6°) = −0.42 km
Dy3 = d3(sin q3) = (1.2 km)(sin 114.6°) = 1.1 km
∆xtot = ∆x1 + ∆x2 + ∆x3 = −4.1 km + 0 km + (−0.42 km) = −4.5 km
∆ytot = ∆y1 + ∆y2 + ∆y3 = 0 km + 17.3 km + 1.1 km = 18.4 km
d = (∆
xtot
)2苶
+苶(∆
ytot
)2 = (−
)2苶
+苶(18
)2
苶苶
苶苶
苶苶
苶苶
苶4.
苶5苶km
苶苶
苶.4
苶苶km
苶苶
= 2.
01苶km
苶0苶×苶苶1苶
苶2苶+苶苶339
苶苶km
苶2苶 = 35
苶9苶km
苶2苶
d = 18.9 km
冢 冣
冢
冣
∆ytot
18.4 km
= tan−1  = −76° = 76° north of west
q = tan−1 
∆xtot
−4.5 km
3. d1 = 850 m
∆x1 = d1(cos q1) = (850 m)(cos 0.0°) = 850 m
q1 = 0.0°
∆y1 = d1(sin q1) = (850 m)(sin 0.0°) = 0 m
d2 = 640 m
∆x2 = d2(cos q2) = (640 m)(cos 36°) = 520 m
q2 = 36°
∆y2 = d2(sin q2) = (640 m)(sin 36°) = 380 m
∆xtot = ∆x1 + ∆x2 = 850 m + 520 m = 1370 m
∆ytot = ∆y1 + ∆y2 = 0 m + 380 m = 380 m
2
2
d = (∆x
+ (∆y苶
m)2 + 苶
(380 m苶
)2 = 1.88
×苶
106 m2苶
+ 1.4 ×苶
105 m2苶
苶
苶苶
苶
tot)苶
tot) = (1370
= 2.02
×苶
106 m2
苶
Copyright © by Holt, Rinehart and Winston. All rights reserved.
d = 1420 m
冢 冣
冢
冣
∆ytot
380 m
q = tan−1 
= tan−1 
∆xtot
1370 m
q = 16° to the side of the initial displacement
V
Section Five—Problem Bank
V Ch. 3–5
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Givens
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Solutions
4. d1 = 2.00 × 102 m
q1 = 0.0°
∆x1 = d1(cos q1) = (2.00 × 102 m)(cos 0.02) = 2.0 × 102 m
∆y1 = d1(sin q1) = (2.00 × 102 m)(sin 0.0°) = 0 m
2
d2 = 3.00 × 10 m
∆x2 = d2(cos q2) = (3.00 × 102 m)(cos 3.0°) = 3.0 × 102 m
q2 = 3.0°
∆y2 = d2(sin q2) = (3.00 × 102 m)(sin 3.0°) = 16 m
2
d3 = 2.00 × 10 m
q3 = 8.8°
∆x3 = d3(cos q3) = (2.00 × 102 m)(cos 8.8°) = 2.0 × 102 m
∆y3 = d3(sin q3) = (2.00 × 102 m)(sin 8.8°) = 31 m
∆xtot = ∆x1 + ∆x2 + ∆x3 = 2.0 × 102 m + 3.0 × 102 m + 2.0 × 102 m = 7.0 × 102 m
∆ytot = ∆y1 + ∆y2 + ∆y3 = 0 m + 16 m + 31 m = 47 m
3 2
d = (∆
xt苶
)2苶
+苶(∆
yt苶
)2 = (7
02苶
m苶
)2苶
+苶(47
m苶
)2 = 4.
05苶
m2苶+
0苶
m苶
苶苶
苶苶
苶.0
苶苶×苶1苶
苶苶
苶9苶×苶苶1苶
苶苶2.2
苶苶×苶1苶苶
ot 苶
ot 苶
= 4.
05苶
m2苶
苶9苶×苶苶1苶
d = 7.0 × 102 m
冢 冣
冢
冣
∆ytot
47 m
q = tan−1 
= tan−1 
∆xtot
7.0 × 102 m
q = 3.8° above the horizontal
5. d1 = 46 km
q1 = 15° south of east
= −15°
d2 = 22 km
∆x1 = d1(cos q1) = (46 km)[cos(−15°)] = 44 km
∆y1 = d1(sin q1) = (46 km)[sin(−15°)] = −12 km
∆x2 = d2(cos q2) = (22 km)[cos(−77°)] = 4.9 km
∆y2 = d2(sin q2) = (22 km)[sin(−77°)] = −21 km
q2 = 13° east of south
= −77°
∆x3 = d3(cos q3) = (14 km)[cos(−104°)] = −3.4 km
d3 = 14 km
∆y3 = d3(sin q3) = (14 km)[sin(−104°)] = −14 km
q3 = 14° west of south
= −90.0° − 14° = −104°
∆xtot = ∆x1 + ∆x2 + ∆x3 = 44 km + 4.9 km + (−3.4 km) = 46 km
∆ytot = ∆y1 + ∆y2 + ∆y3 = −12 km + (−21 km) + (−14 km) = −47 km
= 4.
苶3苶×苶苶10苶3苶km
苶2苶
d = 66 km
冢 冣
冢
冣
∆ytot
−47 km
= tan−1  = −46°
q = tan−1 
∆xtot
46 km
q = 46° south of east
V
V Ch. 3–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
d = (∆
xtot
)2苶
+苶(∆
ytot
)2 = (4
)2苶
+苶(−
)2 = 2.
03苶km
03苶km
苶苶
苶苶
苶苶
苶苶
苶6苶km
苶苶
苶47
苶苶km
苶苶
苶1苶×苶苶1苶
苶2苶+苶苶2.2
苶苶×苶1苶
苶2苶
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Lesson
Givens
Solutions
6. d1 = 6.3 × 108 km
q1 = 0.0°
∆x1 = d1(cos q1) = (6.3 × 108 km)(cos 0.0°) = 6.3 × 108 km
∆y1 = d1 (sin q1) = (6.3 × 108 km)(sin 0.0°) = 0 km
8
d2 = 9.4 × 10 km
∆x2 = d2(cos q2) = (9.4 × 108 km)(cos 68°) = 3.5 × 108 km
q2 = 68°
∆y2 = d2 (sin q2) = (9.4 × 108 km)(sin 68°) = 8.7 × 108 km
9
d3 = 3.4 × 10 km
q3 = 94° + 68° = 162°
∆x3 = d3 (cos q3) = (3.4 × 109 km)(cos 162°) = −3.2 × 109 km
∆y3 = d3 (sin q3) = (3.4 × 109 km)(sin 162°) = 1.1 × 109 km
∆xtot = ∆x1 + ∆x2 + ∆x3 = 6.3 × 108 km + 3.5 × 108 km + (−3.2 × 109 km) = −2.2 × 109 km
∆ytot = ∆y1 + ∆y2 + ∆y3 = 0 km + 8.7 × 103 km + 1.1 × 109 km = 2.0 × 109 km
2
d = (x
yt苶
)2 = (−
09苶km
)2苶
+苶(2.
09苶km
)2
苶苶to苶
苶苶(苶
苶2.
苶2苶×苶苶1苶
苶苶
苶0苶×苶苶1苶
苶苶
t)苶+
ot苶
= 4.
018苶km
018苶km
018苶km
苶8苶×苶苶1苶
苶2苶+苶苶4.0
苶苶×苶1苶
苶2苶 = 8.
苶8苶×苶苶1苶
苶2苶
d = 3.0 × 109 km
冢 冣
冢
冣
∆ytot
2.0 × 109 km
= tan−1 
= − 42°
q = tan−1 
∆xtot
−2.2 × 109 km
180.0° − 42° = 138°
q = 138° from the probe’s initial direction
7. di = 2.50 × 103 m
∆x1 = d1(cos q1) = (2.50 × 103 m)(cos 58.5°) = 1310 m
q1 = 58.5° north of east
∆y1 = d1(sin q1) = (2.50 × 103 m)(sin 58.5°) = 2130 m
d2 = 375 m
∆x2 = d2(cos q2) = (375 m)(cos 21.8°) = 348 m
q2 = 21.8° north of east
∆y2 = d2(sin q2) = (375 m)(sin 21.8°) = 139 m
d3 = 875 m
q3 = 21.5° east of north
∆x3 = d3(sin q3) = (875 m)(sin 21.5°) = 321 m
∆y3 = d3(cos q3) = (875 m)(cos 21.5°) = 814 m
∆xtot = ∆x1 + ∆x2 + ∆x3 = 1310 m + 348 m + 321 m = 1.98 × 103 m
∆ytot = ∆y1 + ∆y2 + ∆y3 = 2130 m + 139 m + 814 m = 3.08 × 103 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
d = (∆
xt苶
)2苶
+苶(∆
yt苶
)2 = (1
03苶
m苶
)2苶
+苶(3.
03苶
m苶
)2
苶苶
苶苶
苶.9
苶8苶×苶苶1苶
苶08
苶苶×苶1苶
ot苶
ot苶
= 3.
06苶
m2苶+
06苶
m2苶 = 13
06苶
m2苶
苶92
苶苶×苶1苶
苶苶9.4
苶9苶×苶苶1苶
苶.4
苶1苶×苶苶1苶
d = 3.66 × 103 m
冢 冣
冢
冣
3.08 × 103 m
∆ytot
= tan−1 
q = tan−1 
∆xtot
1.98 × 103 m
q = 57.3° north of east
V
Section Five—Problem Bank
V Ch. 3–7
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Givens
Solutions
8. d1 = 5.0 km
q1 = 36.9° south of east
= −36.9°
d2 = 1.5 km
∆x1 = d1(cos q1) = (5.0 km)[cos(−36.9°)] = 4.0 km
∆y1 = d1(sin q1) = (5.0 km)[sin(−36.9°)] = −3.0 km
∆x2 = d2(cos q2) = (1.5 km)[cos(−90.0°)] = 0 km
∆y2 = d2(sin q2) = (1.5 km)[sin(−90.0°)] = −1.5 km
q2 = 90.0° due south
= −90.0°
∆x3 = d3(cos q3) = (8.5 km)[cos(−42.2°)] = 6.3 km
d3 = 8.5 km
∆y3 = d3(sin q3) = (8.5 km)[sin(−42.2°)] = −5.7 km
q3 = 42.2° south of east
= −42.2°
d4 = 0.8 km
q4 = 0° (due east)
∆x4 = d4(cos q4) = (0.8 km)(cos 0°) = 0.8 km
∆y4 = d4(sin q4) = (0.8 km)(sin 0°) = 0 km
∆xtot = ∆x1 + ∆x2 + ∆x3 + ∆x4 = 4.0 km + 0 km + 6.3 km + 0.8 km = 11.1 km
∆ytot = ∆y1 + ∆y2 + ∆y3 + ∆y4 = (− 3.0 km) + (−1.5 km) + (−5.7 km) + 0 km = −10.2 km
d = (∆
xtot
)2苶
+苶(∆
ytot
)2 = (1
)2苶
+苶(−
)2 = 12
苶苶
苶苶
苶苶
苶苶
苶1.
苶1苶km
苶苶
苶10
苶.2
苶苶km
苶苶
苶3苶km
苶2苶+苶苶104
苶苶km
苶2苶
= 22
苶7苶km
苶2苶
d = 15.1 km
冢 冣
冢
冣
∆ytot
−10.2 km
q = tan−1 
= tan−1  = −42.6°
∆xtot
11.1 km
q = 42.6° south of east
q1 = 45.0° west of north
= 90.0° + 45.0° = 135.0°
d2 = 1.98 m
q2 = 45.0° east of north
= 45.0°
∆x1 = d1(cos q1) = (1.41 m)(cos 135.0°) = −0.997 m
∆y1 = d1(sin q1) = (1.41 m)(sin 135.0°) = 0.997 m
∆x2 = d2(cos q 2) = (1.98 m)(cos 45.0°) = 1.40 m
∆y2 = d2(sin q 2) = (1.98 m)(sin 45.0°) = 1.40 m
∆x3 = d3(cos q3) = (0.42 m)(cos 135.0°) = −0.30 m
d3 = 0.42 m
∆y3 = d3(sin q3) = (0.42 m)(sin 135.0°) = 0.30 m
q3 = 45.0° west of north
= 135.0°
∆x4 = d4(cos q4) = (1.56 m)(cos 225.0°) = −1.10 m
d4 = 1.56 m
∆y4 = d4(sin q4) = (1.56 m)(sin 225.0°) = −1.10 m
q4 = 45.0° south of west
= 180.0° + 45.0° = 225.0°
∆xtot = ∆x1 + ∆x2 + ∆x3 + ∆x4 = (−0.997 m) + 1.40 m + (−0.30 m) + (−1.10 m)
= −0.997 m
∆ytot = ∆y1 + ∆y2 + ∆y3 + ∆y4 = 0.997 m + 1.40 m + 0.30 m + (−1.10 m) = 1.60 m
d = (∆
xtot
)2苶
+苶(∆
ytot
)2苶
+苶(1.
m苶
)2 = 0.
苶苶
苶苶
苶苶
苶2苶 = (−
苶0.
苶99
苶7苶m
苶苶
苶60
苶苶
苶99
苶4苶m
苶2苶+苶苶2.5
苶6苶m
苶2苶
= 3.
m2苶
苶55
苶苶
d = 1.88 m
冢 冣
冢
冣
∆ytot
1.60 m
= tan−1  = −58.1°
q = tan−1 
∆xtot
−0.997 m
q = 58.1° north of west
V
V Ch. 3–8
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
9. d1 = 1.41 m
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Givens
Solutions
10. d1 = 790 km
∆x1 = d1(cos q1) = (790 km)(cos 162°) = −750 km
q1 = 18° north of west
180.0° − 18° = 162°
d2 = 150 km
q2 = 180.0° due west
∆y1 = d1(sin q1) = (790 km)(sin 162°) = 24 km
∆x2 = d2(cos q2) = (150 km)(cos 180.0°) = −150 km
∆y2 = d2(sin q2) = (150 km)(sin 180.0°) = 0 km
d3 = 470 km
∆x3 = d3(cos q3) = (470 km)(cos 90.0°) = 0 km
q3 = 90.0° due north
∆y3 = d3(sin q3) = (470 km)(sin 90.0°) = 470 km
d4 = 240 km
∆x4 = d4(cos q4) = (240 km)(cos 75°) = 62 km
q4 = 15° east of north
90.0° − 15° = 75°
∆y4 = d4(sin q4) = (240 km)(sin 75°) = 230 km
∆xtot = ∆x1 + ∆x2 + ∆x3 + ∆x4 = (−750 km) + (−150 km) + 0 km + 62 km = −840 km
∆ytot = ∆y1 + ∆y2 + ∆y3 + ∆y4 = 240 km + 0 km + 470 km + 230 km = 940 km
d = (∆
xtot
)2苶
+苶(∆
ytot
)2 = (−
)2苶
+苶(94
)2 = 7.
05苶km
05苶km
苶苶
苶苶
苶苶
苶苶
苶84
苶0苶km
苶苶
苶0苶km
苶苶
苶1苶×苶苶1苶
苶2苶+苶苶8.8
苶苶×苶1苶
苶2苶
= 15
05苶km
苶.9
苶苶×苶1苶
苶2苶
d = 1260 km
冢 冣
冢
冣
∆ytot
940 km
= tan−1  = −48°
q = tan−1 
∆xtot
−840 km
q = 48° north of west
Additional Practice D
∆x
∆t = 
vx
1. vx = 430 m/s
∆x = 4020 m
1
2
ay = −g = −9.81 m/s
∆y = 2ay (∆t)2
2
冢 冣
冢
2
冣
∆x
4020 m
1
1
∆y = 2ay  = 2(−9.81 m/s2)  = −430 m
vx
430 m/s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
height of ridge = 430 m
∆x
∆t = 
vx
2. ∆x = 101 m
vx = 14.25 m/s
1
ay = −g = −9.81 m/s2
∆y = 2ay (∆t)2
2
冢 冣
冢
2
冣
∆x
101 m
1
1
∆y = 2ay  = 2(−9.81 m/s2)  = 246 m
vx
14.25 m/s
height of building = 246 m
3. vx = 1.30 × 102 km/h
∆x = 135 m
∆x
∆t = 
vx
1
2
ay = −g = −9.81 m/s
∆y = 2ay (∆t)2
2
冢 冣
冢
2
冣冢
∆x
135 m
1
1
∆y = 2ay  = 2(−9.81 m/s2) 
vx
1.30 × 102 km/h
2
冣
3600 s/h

 = −68.6 m
103 m/km
airship’s altitude = 68.6 m
V
Section Five—Problem Bank
V Ch. 3–9