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NAME ______________________________________ DATE _______________ CLASS ____________________
Two-Dimensional Motion and Vectors
Problem C
ADDING VECTORS ALGEBRAICALLY
PROBLEM
The record for the longest nonstop closed-circuit flight by a model airplane was set in Italy in 1986. The plane flew a total distance of 1239 km.
Assume that at some point the plane traveled 1.25 × 103 m to the east, then
1.25 × 103 m to the north, and finally 1.00 × 103 m to the southeast. Calculate the total displacement for this portion of the flight.
SOLUTION
1. DEFINE
Given:
d1 = 1.25 × 103 m
d2 = 1.25 × 103 m
d3 = 1.00 × 103 m
Unknown:
∆xtot = ? ∆ytot = ?
d =?
q=?
Diagram:
d3 = 1.00 × 103 m
d2 = 1.25 × 103 m
45.0°
dtot
∆ytot
d1 = 1.25 × 103 m
∆ xtot
2. PLAN
N
Choose the equation(s) or situation: Orient the displacements with respect to
the x-axis of the coordinate system.
Copyright © by Holt, Rinehart and Winston. Allrights reserved.
q1 = 0.00°
q2 = 90.0°
q3 = −45.0°
Use this information to calculate the components of the total displacement along
the x-axis and the y-axis.
∆xtot = ∆x1 + ∆x2 + ∆x3
= d1(cos q1) + d2 (cos q2) + d3 (cos q3)
∆ytot = ∆y1 + ∆y2 + ∆y3
= d1(sin q1) + d2 (sin q2) + d3 (sin q3)
3. CALCULATE
Use the components of the total displacement, the Pythagorean theorem, and the
tangent function to calculate the total displacement.
∆ytot
d= (∆
xtot
)2
+(
∆
ytot
q = tan−1 )2
∆xtot
Substitute the values into the equation(s) and solve:
∆xtot = (1.25 × 103 m)(cos 0°) + (1.25 × 103 m)(cos 90.0°)
+(1.00 × 103 m)[cos (−45.0°)]
= 1.25 × 103 m + 7.07 × 102 m
= 1.96 × 103 m
∆ytot = (1.25 × 103 m)(sin 0°) + (1.25 × 103 m)(sin 90.0°)
+(1.00 × 103 m)[sin (−45.0°)]
= 1.25 × 103 m + 7.07 × 102 m
= 0.543 × 103 m
d = (1
m)2
+(0.
3×
m)2
.9
6×103
54
103
Problem C
21
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NAME ______________________________________ DATE _______________ CLASS ____________________
d = 3.
m2+
5×
m2 = 4.
m2
84
×106
2.9
105
14
×106
d = 2.03 × 103 m
0.543 × 103 m
q = tan−1 
1.96 × 103 m
q = 15.5° north of east
4. EVALUATE
The magnitude of the total displacement is slightly larger than that of the total
displacement in the eastern direction alone.
ADDITIONAL PRACTICE
1. For six weeks in 1992, Akira Matsushima, from Japan, rode a unicycle
more than 3000 mi across the United States. Suppose Matsushima is riding through a city. If he travels 250.0 m east on one street, then turns
counterclockwise through a 120.0° angle and proceeds 125.0 m northwest along a diagonal street, what is his resultant displacement?
3. Magnor Mydland of Norway constructed a motorcycle with a wheelbase
of about 12 cm. The tiny vehicle could be ridden at a maximum speed
11.6 km/h. Suppose this motorcycle travels in the directions d1 and d2
shown in the figure below. Calculate d1 and d2, and determine how long
it takes the motorcycle to reach a net displacement of 2.0 × 102 m to the
right?
4. The fastest propeller-driven aircraft is the Russian TU-95/142, which can
reach a maximum speed of 925 km/h. For this speed, calculate the plane’s
resultant displacement if it travels east for 1.50 h, then turns 135° northwest and travels for 2.00 h.
5. In 1952, the ocean liner United States crossed the Atlantic Ocean in less
than four days, setting the world record for commercial ocean-going vessels. The average speed for the trip was 57.2 km/h. Suppose the ship
moves in a straight line eastward at this speed for 2.50 h. Then, due to a
strong local current, the ship’s course begins to deviate northward by
30.0°, and the ship follows the new course at the same speed for another
1.50 h. Find the resultant displacement for the 4.00 h period.
22
Holt Physics Problem Workbook
Copyright © by Holt, Rinehart and Winston. Allrights reserved.
2. In 1976, the Lockheed SR-71A Blackbird set the record speed for any airplane: 3.53 × 103 km/h. Suppose you observe this plane ascending at this
speed. For 20.0 s, it flies at an angle of 15.0° above the horizontal, then
for another 10.0 s its angle of ascent is increased to 35.0°. Calculate the
plane’s total gain in altitude, its total horizontal displacement, and its resultant displacement.
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Givens
4. d = (8)(4.5 m)
q = 35°
Solutions
∆x = d(cos q) = (8)(4.5 m)(cos 35°) = 29 m
∆y = d(sin q) = (8)(4.5 m)(sin 35°) = 21 m
5. v = 347 km/h
vx = v(cos q) = (347 km/h)(cos 15.0°) = 335 km/h
q = 15.0°
vy = v(sin q) = (347 km/h)(sin 15.0°) = 89.8 km/h
6. v = 372 km/h
∆t = 8.7 s
q = 60.0°
1h
d = v∆t = (372 km/h)  (103 m/km)(8.7 s) = 9.0 × 102 m
3600 s
∆x = d(cos q) = (9.0 × 102 m)(cos 60.0°) = 450 m east
∆y = d(sin q ) = (9.0 × 102 m)(sin 60.0°) = 780 m north
7. d = 14 890 km
q = 25.0°
d 1.489 × 104 km
vavg =  =  = 805 km/h
∆t
18.45 h
∆t = 18.5 h
vx = vavg (cos q) = (805 km/h)(cos 25.0°) = 730 km/h east
vy = vavg (sin q) = (805 km/h)(sin 25.0°) = 340 km/h south
8. vi = 6.0 × 102 km/h
vf = 2.3 × 103 km/h
∆t = 120 s
q = 35° with respect to
horizontal
II
∆v vf − vi
a =  =  
∆t
∆t
1h
(2.3 × 103 km/h − 6.0 × 102 km/h)  (103 m/km)
3600 s
a = 
1.2 × 102 s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1h
(1.7 × 103 km/h)  (103 m/km)
3600 s
a = 
1.2 × 102 s
2
a = 3.9 m/s
ax = a(cos q) = (3.9 m/s2)(cos 35°) = 3.2 m/s2 horizontally
ay = a(sin q) = (3.9 m/s2)(sin 35°) = 2.2 m/s2 vertically
Additional Practice C
1. ∆x1 = 250.0 m
∆x2 = d2 (cos q 2 ) = (125.0 m)(cos 120.0°) = −62.50 m
d2 = 125.0 m
∆y2 = d2 (sin q 2 ) = (125.0 m)(sin 120.0°) = 108.3 m
q 2 = 120.0°
∆xtot = ∆x1 + ∆x2 = 250.0 m − 62.50 m = 187.5 m
∆ytot = ∆y 1 + ∆y2 = 0 m + 108.3 m = 108.3 m
d = (∆
xtot
)2
+(∆
ytot
)2 = (1
m
)2
+(10
)2
87
.5
8.
3m
d = 3.
04
m2+
04
m2 =
51
6×1
1.1
73
×1
4.
m2
68
9×104
d = 216.5 m
∆ytot
108.3 m
= tan−1  = 30.01° north of east
q = tan−1 
∆xtot
187.5 m
Section Two — Problem Workbook Solutions
II Ch. 3–3
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Givens
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Solutions
2. v = 3.53 × 103 km/h
∆t1 = 20.0 s
∆t2 = 10.0 s
q1 = 15.0°
q2 = 35.0°
∆x1 = v∆t1(cos q 1 )
1h
∆x1 = (3.53 × 103 km/h)  (103 m/km)(20.0 s)(cos 15.0°) = 1.89 × 104 m
3600 s
∆y1 = v∆t1(sin q 1 )
1h
∆y1 = (3.53 × 103 km/h)  (103 m/km)(20.0 s)(sin 15.0°) = 5.08 × 103 m
3600 s
∆x2 = v∆t2 (cos q 2 )
1h
∆x2 = (3.53 × 103 km/h)  (103 m/km)(10.0 s)(cos 35.0°) = 8.03 × 103 m
3600 s
∆y2 = v∆t2 (sin q 2 )
1h
∆y2 = (3.53 × 103 km/h)  (103 m/km)(10.0 s)(sin 35.0°) = 5.62 × 103 m
3600 s
∆ytot = ∆y1 + ∆y2 = 5.08 × 103 m + 5.62 × 103 m = 1.07 × 104 m
∆xtot = ∆x1 + ∆x2 = 1.89 × 104 m + 8.03 × 103 m = 2.69 × 104 m
d = (∆
xt
)2
+(∆
yt
)2 = (2
04
m
)2
+(1.
04
m
)2
.6
9×1
07
×1
ot
ot
II
d = 7.
08
m2+
1×
08
m2 = 8.
08
m
24
×1
1.1
1
35
×1
d = 2.89 × 104 m
∆ytot
1.07 × 104 m
q = tan−1  = tan−1 
∆xtot
2.69 × 104 m
q = 21.7° above the horizontal
∆y1 + ∆y2 = 0
q1 = 30.0°
q2 = −45.0°
v = 11.6 km/h
∆y1 = d1(sin q1) = −∆y2 = −d2(sin q2)
sin q
sin(− 45.0°)
d1 = −d2 2 = −d2  = 1.41d2
sin q1
sin 30.0°
∆x1 = d1(cos q1) = (1.41d2)(cos 30.0°) = 1.22d2
∆x2 = d2(cos q2) = d2[cos(−45.0°)] = 0.707d2
∆x1 + ∆x2 = d2(1.22 + 0.707) = 1.93d2 = 2.00 × 102 m
d2 = 104 m
d1 = (1.41)d2 = (1.41)(104 m) = 147 m
1h
v = 11.6 km/h = (11.6 km/h)  (103 m/km) = 3.22 m/s
3600 s
104 m
d
∆t =  =  = 32.3 s
v 3.22 m/s
147 m
d
∆t1 = 1 =  = 45.7 s
3.22 m/s
v
2
2
∆ttot = ∆t1 + ∆t2 = 45.7 s + 32.3 s = 78.0 s
II Ch. 3–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
3. ∆x1 + ∆x2 = 2.00 × 102 m
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Givens
4. v = 925 km/h
Solutions
d1 = v∆t1 = (925 km/h)(103 m/km)(1.50 h) = 1.39 × 106 m
∆t1 = 1.50 h
d2 = v∆t2 = (925 km/h)(103 m/km)(2.00 h) = 1.85 × 106 m
∆t2 = 2.00 h
∆x1 = d1 = 1.39 × 106 m
q 2 = 135°
∆y1 = 0 m
∆x2 = d2 (cos q 2 ) = (1.85 × 106 m)(cos 135°) = −1.31 × 106 m
∆y2 = d2 (sin q 2 ) = (1.85 × 106 m)(sin 135°) = 1.31 × 106 m
∆xtot = ∆x1 + ∆x2 = 1.39 × 106 m + (− 1.31 × 106 m) = 0.08 × 106 m
∆ytot = ∆y1 + ∆y2 = 0 m + 1.31 × 106 m = 1.31 × 106 m
d = (∆
xtot
m)2
+(1.
m)2
)2+(∆
ytot)2 = (0
.0
8×106
31
×106
2 2
d = 6×
09
m2+
01
m
012m
1
1.7
2×1
= 1.
73
×1
2
d = 1.32 × 106 m = 1.32 × 103 km
∆ytot
1.31 × 106 m
q = tan−1 
= tan−1 
= 86.5° = 90.0° − 3.5°
∆xtot
0.08 × 106 m
II
q = 3.5° east of north
5. v = 57.2 km/h
d1 = v∆t1 = (57.2 km/h)(2.50 h) = 143 km
∆t1 = 2.50 h
d2 = v∆t2 = (57.2 km/h)(1.50 h) = 85.8 km
∆t2 = 1.50 h
∆tot = d1 + d2(cos q2) = 143 km + (85.8 km)(cos 30.0°) = 143 km + 74.3 km = 217 km
θ2 = 30.0°
∆ytot = d2(sin q2) = (85.8 km)(sin 30.0°) = 42.9 km
xtot
)2
+(∆
ytot
)2 = (2
)2
+(42
d = (∆
17
km
.9
km
)
d = 4.
04km
03km
04km
71
×1
2+1.8
4×1
2 = 4.
89
×1
2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
d = 221 km
42.9 km
∆ytot
q = tan−1  = tan−1  = 11.2° north of east
217 km
∆xtot
Additional Practice D
1. vx = 9.37 m/s
∆y = −2.00 m
ay = −g = −9.81 m/s2
∆t =
2∆y ∆ x
 = 

v
a
x
y
∆x = vx
2∆y
(2)(−2.00 m)
 = (9.37 m/s) 
 = 5.98 m

a
(−9.81
m/s )
2
y
The river is 5.98 m wide.
2. ∆x = 7.32 km
∆y = −8848 m
ay = −g = −9.81 m/s2
∆t =
2∆y ∆ x
 = 

v
a
y
vx =
x
2
(−9.81 m/s )
 ∆x =   (7.32 × 10 m) =

2∆y
(2)(−8 848 m)
ay
3
172 m/s
No. The arrow must have a horizontal speed of 172 m/s, which is much greater than
100 m/s.
Section Two — Problem Workbook Solutions
II Ch. 3–5