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NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 3F
RELATIVE VELOCITY
PROBLEM
A polar bear swims 2.60 m/s south relative to the water. The bear is swimming against a current that moves 0.78 m/s at an angle of 40.0° north of
west, relative to Earth. How long will it take the polar bear to reach the
shore, which is 5.50 km to the south?
1. DEFINE
Given:
vbc = 2.60 m/s due south (velocity of the bear, b, with respect to
the current, c)
vce = 0.78 m/s at 40.0° north of west (velocity of the current, c,
with respect to Earth, e)
∆y = 5.50 km, south
Unknown:
∆t = ?
Diagram:
N
vbe
θce = 40.0°
2. PLAN
vbc
vce
Choose the equation(s) or situation: To find vbe , write the equation so that the
subscripts of the vectors on the right begin with b and end with e.
vbe = vbc + vce
Because vectors vbc and vce are not perpendicular, their x and y components
must be calculated. Aligning the positive y axis with north and treating west as
the positive x direction for convenience, the following equations apply for the
magnitude of the components of vbe .
vx,be = vx,bc + vx,ce = vx,ce = vce (cos q ce)
vy,be = vy,bc + vy,ce = −vbc + vce (sin q ce)
From these components the magnitude and direction of vbe could be found
from the Pythagorean theorem and the tangent function, respectively. However,
only the component vy,be is needed to calculate the time required for the bear to
swim in the negative y direction.
−∆y
∆t = vy,be
Ch. 3–16
Holt Physics Problem Bank
Copyright © by Holt, Rinehart and Winston. All rights reserved.
SOLUTION
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NAME ______________________________________ DATE _______________ CLASS ____________________
Rearrange the equation(s) to isolate the unknown(s):
−∆y
−∆y
∆t = = vy,be [−vbc + vce (sin q ce)]
3. CALCULATE
Substitute the values into the equation(s) and solve:
−5.50 km
∆t = (−2.60 m/s + 0.78 m/s)(sin 40.0°)]
−5.50 km
−5.50 × 103 m
= = −2.10 m/s
(−2.60 m/s + 0.50 m/s)
∆t = 2.62 × 103 s, or 43 min 40 s
4. EVALUATE
Without the current, the polar bear would arrive about 500 s or 8.3 min sooner.
The 500 s delay is about one fourth (25%) of the bear’s swimming time without
the current. This proportion is equal to the ratio of the current’s northern component to the bear’s velocity to the south.
ADDITIONAL PRACTICE
1. A bird flies directly into a wind. If the bird’s forward speed relative to the
wind is 58.0 km/h and the wind’s speed in the opposite direction is
55.0 km/h, relative to Earth, how long will it take the bird to fly 1.4 km?
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2. A moving walkway at an airport has a velocity of 1.50 m/s to the west. A
man rushing to catch his flight runs down the walkway with a velocity of
4.20 m/s to the west relative to the walkway. If the walkway if 8.50 ×
102 m long, how much time does the man save by running on the walkway as opposed to running on a non-moving surface?
3. The greatest average speed for a race car in the Daytona 500 is 286 km/h,
which was achieved in 1980. Suppose a race car moving at this speed is in
second place, being 0.750 km behind a car that is moving at a speed of
252 km/h. How long will it take the second-place car to catch up to the
first-place car?
4. A mosquito can fly with a speed of 1.10 m/s with respect to the air. Suppose a mosquito flies east at this speed across a swamp. The mosquito is
flying into a breeze that has a velocity of 5.0 km/h with respect to Earth
and moves 35° west of south. If the swamp is 540 m across, how long will
it take the mosquito to cross the swamp?
5. A glider descends with a velocity relative to the air of 150 km/h at an
angle of 7.0° below the horizontal. Suppose that the glider encounters an
updraft with a velocity relative to Earth of 15 km/h upward. How long
will it take the glider to reach the ground if it encounters the updraft at
166 m? How long would it take for the glider to land without the updraft?
6. A flare gun is mounted on an automobile and fired perpendicular to the
car’s motion. The car’s velocity with respect to Earth is 145 km/h to the
north. The flare’s velocity with respect to the car is 87 km/h to the west.
What are the components of the flare’s displacement with respect to
Earth 0.45 s after the flare is launched?
Problem 3F
Ch. 3–17
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7. An airship moving north at 55.0 km/h with respect to the air encounters a wind from 17.0° north of west. If the wind’s speed with respect to
Earth is 40.0 km/h, what is the airship’s velocity with respect to Earth?
8. How far to the north and west does the airship in problem 7 travel after
15.0 minutes?
9. A torpedo fired at an anchored target moves against a current. Suppose
the torpedo’s velocity with respect to the current is 51 km/h east, and
the current’s velocity with respect to the target is 4.0 km/h south. If the
torpedo hits the target in 14 s, how far away is the target from the point
where the torpedo is launched? How far north of the target must the
torpedo be launched in order to hit the target?
Copyright © by Holt, Rinehart and Winston. All rights reserved.
10. A sailboat travels south with a speed of 12.0 km/h with respect to the
water. Suppose the boat encounters a current that has a velocity with
respect to Earth of 4.0 km/h at 15.0° south of east. What is the sailboat’s resultant velocity with respect to Earth?
Ch. 3–18
Holt Physics Problem Bank
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Givens
Solutions
10. ∆x1 = 0.46 m
∆xtot = ∆x1 + ∆x2
∆x
∆t = 
vi(cos q)
∆x2 = 4.00 m
∆xi + ∆x2 1 ∆x1 + ∆x2
1
 − 2g 
∆y = vi(sin q)∆t − 2g∆t 2 = vi(sin q) 
vi(cos q)
vi(cos q)
∆y = − 0.35 m
g(∆x1 + ∆x2 )2

∆y = (∆x1 + ∆x2)(tan q) − 
2 vi2(cos q)2
g = 9.81 m/s2
vi =
vi =
vi =
vi =
2
q = 41.0°
g(∆x1 + ∆x2)2

2
2(cos q) [(∆x1 + ∆x2)(tan q) − ∆y]
(9.81 m/s2)(0.46 m + 4.00 m)2

2
(2)(cos 41.0°) [(0.46 m + 4.00 m)(tan 41.0°) − (−0.35 m)]
(9.81 m/s2)(4.46 m)2

(2)(cos 41.0°)2(3.88 m + 0.35 m)
(9.81 m/s2)(4.46 m)2

(2)(cos 41.0°)2 (4.23 m)
vi = 6.36 m/s
Additional Practice 3F
1. vbw = 58.0 km/h, forward
= +58.0 km/h
vwe = 55.0 km/h, backward
= −55.0 km/h
vbe = vbw + vwe = + 58.0 km/h + (−55.0 km/h) = +3.0 km/h
∆x 1.4 km
∆t =  = 
vbe 3.0 km/h
∆t = 0.47 h = 28 min
∆x = 1.4 km
vmw = 4.20 m/s, west
2
∆x = 8.50 × 10 m
vme = vmw + vwe = 4.20 m/s + 1.50 m/s = 5.70 m/s, west
time of travel with walkway:
∆x 8.50 × 102 m
∆t1 =  =  = 149 s
vme
5.70 m/s
time of travel without walkway:
∆x 8.50 × 102 m
∆t2 =  =  = 202 s
vmw
4.20 m/s
time saved = ∆t2 − ∆t1 = 202 s − 149 s = 53 s
3. v1e = 286 km/h, forward
v12 + v2e = v1e
v2e = 252 km/h, forward
v12 = v1e − v2e
∆x = 0.750 km
v12 = v1e − v2e = 286 km/h − 252 km/h = 34 km/h
∆x 0.750 km
∆t =  −  = 2.2 × 10−2 h
v12 34 km/h
3600 s
∆t = (2.2 × 10−2 h)  = 79 s
1h
V
V Ch. 3–14
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2. vwe = 1.50 m/s, west
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Givens
4. vma = 1.10 m/s, east
vae = 5.0 km/h at 35°
west of south
∆x = 540 m
Solutions
vme = vma + vae
Find the mosquito’s speed with respect to Earth in the x direction.
vx, me = vx, ma + vx, ae = vma + vae(cos qae)
qae = –90.0° − 35° = −125°
1h
vx, me = 1.10 m/s + (5.0 km/h)(103 m/km)  [cos(−125°)] = 1.10 m/s
3600 s
+ (−0.80 m/s) = 0.30 m/s
∆x
540 m
∆t =  = 
vx, me 0.30 m/s
∆t = 1800 s = 3.0 × 101 min
5. vga = 150 km/h at 7.0°
below horizontal
vae = 15 km/h upward
∆y = −165 m
vge = vga + vae
Find the glider’s speed with respect to Earth in the y (vertical) direction.
vy, ge = vga + vy, ae = vga(sin qga) + vae
qga = −7.0°
vy, ge = (150 km/h)[sin(−7.0°)] + 15 km/h = −18 km/h + 15 km/h = −3 km/h
Time of descent with updraft:
−166 m
∆y
∆t =  = 
vy, ge (−3 km/h)(103 m/km)(1 h/3600 s)
∆t = 200 s
Time of descent without updraft:
−166 m
∆y
∆t′ =  = 
vy, ga (−18 km/h)(103 m/km)(1h/3600 s)
∆t′ = 33 s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
6. vfc = 87 km/h, west
vfe = vfc + vce
vce = 145 km/h, north
vx, fe = vx, fc + vx, ce = vfc = −87 km/h
∆t = 0.45 s
vy, fe = vy, fc + vy, ce = vce = +145 km/h
∆x = vx, fe ∆t = (−87 km/h)(103 m/km)(1 h/3600 s)(0.45 s) = −11 m
∆x = 11m, west
∆y = vy, fe ∆t = (145 km/h)(103 m/km)(1 h/3600 s)(0.45 s)
∆y = 18 m, north
V
Section Five—Problem Bank
V Ch. 3–15
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Solutions
7. vaw = 55.0 km/h, north
vwe = 40.0 km/h at 17.0°
north of west
vae = vaw + vwe
vx, ae = vx, aw + vx, we = vwe (cos qwe)
vy, ae = vy, aw + vy, we = vaw + vwe(sin qwe)
qwe = 180.0° − 17.0° = 163.0°
vx, ae = (40.0 km/h)(cos 163.0°) = −38.3 km/h
vy, ae = 55.0 km/h + (40.0 km/h)(sin 163.0°) = 55.0 km/h + 11.7 km/h = 66.7 km/h
2
2
vy
)2
+(66
)2
vae = vx
(
38
.3
km
/h
.7
km
/h
, a
e)+
, a
e0 = (−
vae = 1.
47
×103km
2/h2+
4.4
5×103km
2/h2 = 5.
92
×103km
2/h2
vae = 76.9 km/h
66.7 km/h
vy,ae
q = tan−1  = tan−1  = −60.1°
−38.3 km/h
vx,ae
q = 60.1° west of north
8. vae = 76.9 km/h at 29.9°
west of north
∆t = 15.0 min
∆x = vae(cos qae)∆t
∆y = vae(sin qae)∆t
qae = 90.0° + 29.9° = 119.9°
∆x = (76.9 km/h)(cos 119.9°)(15.0 min)(1 h/60 min) = −9.58 km
∆y = (76.9 km/h)(sin 119.9°)(15.0 min)(1 h/60 min) = 16.7 km
∆x =
9.58 km, west
∆y = 16.7 km, north
9. vtc = 51 km/h, east
vte = vtc + vce
vce = 4.0 km/h, south
vx, te = vx, tc + vx, ce = vtc = 51 km/h
∆t = 14 s
vy, te = vy, tc + vy, ce = vce = −4.0 km/h
the target is 2.0 × 102 m away
∆y = vy, te ∆t = (4.0 km/h)(103 m/km)(1 h/3600 s)(14 s) = 16 m
the torpedo must be fired 16 m north of the target
V
V Ch. 3–16
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆x = vx, tc∆t = (51 km/h)(103 m/km)(1 h/3600 s)(14 s) = 2.0 × 102 m
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Givens
Solutions
10. vbw = 12.0 km/h, south
vbe = vbw + vwe
vwe = 4.0 km/h at 15.0°
south of east
vx, be = vx, bw + vx, we = vwe(cos qwe)
vy, be = vy, bw + vy, we = vbw + vwe(sin qwe)
qwe = −15.0°
vx, be = (4.0 km/h)[cos(−15.0°)] = 3.9 km/h
vy, be = (−12.0 km/h) + (4.0 km/h)[sin(−15.0°)] = (−12.0 km/h) + (−1.0 km/h)
= −13.0 km/h
2
vbe = (v
)2 = (3
)2
+(−
)2
x,b
v
.9
km
/h
13
.0
km
/h
e)+
y, be
vbe = 15
km
2/h2+169
km
2/h2 = 18
4km
2/h2
vbe = 13.6 km/h
vy, be
−13.0 km/h
q = tan−1  = tan−1  = −73°
vx, be
3.9 km/h
Copyright © by Holt, Rinehart and Winston. All rights reserved.
q = 73° south of east
V
Section Five—Problem Bank
V Ch. 3–17