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of the velocity vSW of the swimmer relative to the
water and the velocity v WG of the water relative
to the ground as shown at the right:
vSG = vSW + v WG . 137 vWG =
0.91 m/s vSW =
1.4 m/s Direction of
current vSG The component of vSG that is parallel to the
width of the river determines how fast the
swimmer is moving across the river; this parallel
component is vSW . The time for the swimmer to cross the river is equal to the width of the
river divided by the magnitude of this velocity component.
The component of vSG that is parallel to the direction of the current determines how far the
swimmer is carried down stream; this component is vWG . Since the motion occurs with
constant velocity, the distance that the swimmer is carried downstream while crossing the
river is equal to the magnitude of vWG multiplied by the time it takes for the swimmer to
cross the river.
SOLUTION
a. The time t for the swimmer to cross the river is t= width 2.8 × 103 m
=
= 2.0 ×103 s
vSW
1.4 m/s b. The distance x that the swimmer is carried downstream while crossing the river is
x = vWG t = (0.91 m/s)(2.0 × 103 s) = 1.8 × 103 m 54. REASONING There are three velocities involved:
v N B = velocity of Neil relative to Barbara
v Ν G = velocity of Neil relative to the Ground (3.2 m/s, due west)
v BG = velocity of Barbara relative to the Ground (4.0 m/s, due south) Ordering the vectors vNB , vNG and vGB by their subscripts in the manner discussed in
Section 3.4 of the text, we see that v NB = v NG + v GB . Note that this ordering involves v GB ,
the velocity of the ground relative to Barbara. According to the discussion in Section 3.4, 138 KINEMATICS IN TWO DIMENSIONS v GB is related to v BG by v GB = − v BG . Thus, the three vectors listed above are related by v NB = v NG + ( − v BG ) . The drawing shows this relationship. North Since the three vectors form a right triangle, the
magnitudes of the vectors are related by the
Pythagorean theorem, which we will use to obtain
Neil’s speed relative to Barbara. Trigonometry can
be used to determine the angle θ.
SOLUTION Using the Pythagorean theorem, we
find that the speed vNB of Neil relative to Barbara is
2
v NB = v NG + ( − vBG ) =
2 v NB − vBG θ West ( 3.2 m/s )2 + ( −4.0 m/s )2 v NG
= 5.1 m/s The angle θ can be found from trigonometry: v NG −1 3.2 m/s = cos = 51° ( north of west ) 5.1 m/s v NB θ = cos −1 55. REASONING As indicated by Equation 3.7, the velocity of the student relative to the
ground is
vSG = v SE + v EG
where vSE is the velocity of the student relative to the escalator and vEG is the velocity of
the escalator relative to the ground. Thus, we have
vSE = v SG – v EG
SOLUTION Taking the direction in which the student runs as the positive direction, we
find that
30.0 m
vSE = vSG – vEG =
– ( –1.8 m/s ) = 4.5 m/s
11 s The student must exceed this speed of 4.5 m/s to beat the record. Chapter 3 Problems 139 56. REASONING There are three velocities involved:
v TG = the velocity of the Truck relative to the Ground
v T P = the velocity of the Truck relative to the Police car
v PG = the velocity of the Police car relative to the Ground Of these three vectors, we know the directions of v P G (north) and v TG (west), and the
magnitudes of v P G (29 m/s) and v T P (48 m/s). Ordering these vectors by their subscripts
in the manner discussed in Section 3.4 of the text, we see that v TG = v TP + v PG . The
triangle formed by this vector sum is a right triangle, because two of the vectors
( v TG and v PG ) are perpendicular to one another. The third vector, v TP , must be the
triangle’s hypotenuse. The magnitudes of the three vectors are thus related to one another by
the Pythagorean theorem, which we will use to obtain the truck’s speed relative to the
ground.
SOLUTION
a. To represent the vector sum
v TG = v TP + v PG graphically, the
vectors being added together, v T P North and v P G , must be drawn tailtohead
(see the drawing). The resultant
vector v TG runs from the tail of
v TP vTG West v PG
v TP to the head of v P G . The direction of v T P shows that, relative
to the police car, the truck is moving
both west and south.
b. As the drawing shows, v T P is the hypotenuse of the vector right triangle. Therefore, the
Pythagorean theorem gives
2
2
2
vTP = vTG + vPG or 2
2
vTG = vTP − vPG The speed of the truck relative to the ground is
2
2
vTG = vTP − vPG = This is equivalent to 85 mph. ( 48 m/s )2 − ( 29 m/s )2 = 38 m/s 140 KINEMATICS IN TWO DIMENSIONS 57. SSM REASONING The velocity v AB of train A relative to train B is the vector sum of the velocity v AG of train A relative to the ground and the velocity v GB of the ground
relative to train B, as indicated by Equation 3.7: v AB = v AG + v GB . The values of v AG and
v BG are given in the statement of the problem. We must also make use of the fact that
v GB = − v BG . SOLUTION
a. Taking east as the positive direction, the velocity of A relative to B is, according to
Equation 3.7, v AB = v AG + v GB = v AG − v BG = (+13 m/s) – (–28 m/s) = +41 m/s
The positive sig...
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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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