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Unformatted text preview: n indicates that the direction of v AB is due east .
b. Similarly, the velocity of B relative to A is
v BA = v BG + v GA = v BG − v AG = (–28 m/s) – (+13 m/s) = –41 m/s
The negative sign indicates that the direction of v BA is due west . 58. REASONING The time it takes for the passenger to walk the distance on the boat is the
distance divided by the passenger’s speed vPB relative to the boat. The time it takes for the
passenger to cover the distance on the water is the distance divided by the passenger’s speed
vPW relative to the water. The passenger’s velocity relative to the boat is given. However, we
need to determine the passenger’s velocity relative to the water.
SOLUTION a. In determining the velocity of the passenger relative to the water, we define
the following symbols: vPW = Passenger’s velocity relative to the water
vPB = Passenger’s velocity relative to the boat
vBW = Boat’s velocity relative to the water
The passenger’s velocity relative to the water is b gb g v PW = v PB + v BW = 1.5 m / s, north + 5.0 m / s, south = 3.5 m / s, south
b. The time it takes for the passenger to walk a distance of 27 m on the boat is Chapter 3 Problems t= 141 27 m
27 m
=
= 18 s
vPB 1.5 m/s c. The time it takes for the passenger to cover a distance of 27 m on the water is
t= 27 m
27 m
=
= 7.7 s
vPW 3.5 m/s ____________________________________________________________________________________________ 59. REASONING Let v HB represent the velocity of the hawk relative to the balloon and v BG
represent the velocity of the balloon relative to the ground. Then, as indicated by
Equation 3.7, the velocity of the hawk relative to the ground is v HG = v HB + v BG . Since the
vectors v HB and v BG are at right angles to each other, the vector addition can be carried
out using the Pythagorean theorem.
SOLUTION Using the drawing at the right, we have
from the Pythagorean theorem, NORTH vHG 2
2
v HG = v HB + v BG θ = (2.0 m / s) 2 + (6.0 m / s) 2 = 6.3 m / s
The angle θ is θ = tan −1 F I = tan F m / s I =
v
2.0
G J G m / sJ
HK
v
6.0
HK
−1 HB vBG vHB
EAST 18 ° , north of east BG 60. REASONING Using subscripts to make the relationships among the three relative
velocities clear, we have the following:
vPG = the velocity of the Plane relative to the Ground (unknown speed, due west)
vPA = the velocity of the Plane relative to the Air (245 m/s, unknown direction)
vAG = the velocity of the Air relative to the Ground (38.0 m/s, due north) Arranging the subscripts as shown in Section 3.4 of the text, we find that the velocity of the
plane relative to the ground is the vector sum of the other two velocities: vPG = vPA + vAG.
This vector sum may be illustrated as follows:
North vPG
vAG θ vPA East 142 KINEMATICS IN TWO DIMENSIONS As the three relative velocity vectors form a right triangle, we will apply trigonometry to
find the direction θ that the pilot should aim the plane relative to due west.
SOLUTION The magnitudes of the two known sides of the vector right triangle are vPA
(245 m/s) and vAG (38.0 m/s) which are, respectively, the hypotenuse and the side opposite
the angle θ (see the drawing). The sine of the angle θ is therefore the ratio of vAG to vPA, so
that the direction the pilot should head the plane is vAG −1 38.0 m/s o = sin = 8.92 , south of west vPA 245 m/s θ = sin −1 61. REASONING AND SOLUTION
by The velocity of the raindrops relative to the train is given
vRT = vRG + vGT where vRG is the velocity of the raindrops relative to the ground and vGT is the velocity of
the ground relative to the train.
Since the train moves horizontally, and the rain falls
vertically, the velocity vectors are related as shown in the
figure at the right. Then vRT 25° vRG vGT = vRG tan θ = (5.0 m/s) (tan 25°) = 2.3 m/s
The train is moving at a speed of 2.3 m/s vGT ____________________________________________________________________________________________ 62. REASONING The three relative velocities in this situation are
vTG = the velocity of the Truck relative to the Ground (unknown)
vTC = the velocity of the Truck relative to the Car (24.0 m/s, 52.0° north of east)
vCG = the velocity of the Car relative to the Ground (16.0 m/s, due north) The velocity of the truck relative to the ground may be expressed as the vector sum of the
other two velocities: vTG = vTC + vCG . Note the fashion in which the “middle” subscripts
on the right side of the equals sign are matched. See the diagram for an illustration of this
vector sum.
Because the vectors do not form a right triangle, we will utilize the component method of
vector addition to determine the eastward and northward components of the resultant vector Chapter 3 Problems vTG. For convenience, we will take east as the +x direction and
north as the +y direction. Once we know vTG,x and vTG,y, we will
use the Pythagorean theorem to calculate the magnitude of the
truck’s velocity relative to the ground, which, from the diagram,
we expect to be larger than vTC = 24.0 m/s.
SOLUTION The vector vCG points...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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