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Unformatted text preview: ONING The meteoroid’s speed is the magnitude of its velocity vector, here described
in terms of two perpendicular components, one directed toward the east and one directed
vertically downward. Let east be the +x direction, and up be the +y direction. Then the
components of the meteoroid’s velocity are vx = +18.3 km/s and vy = −11.5 km/s. The
meteoroid’s speed v is related to these components by the Pythagorean theorem (Equation
2
1.7): v2 = vx + v2 .
y SOLUTION From the Pythagorean theorem,
2
v = vx + v 2 =
y ( +18.3 km/s )2 + ( −11.5 km/s )2 = 21.6 km/s It’s important to note that the negative sign for vy becomes a positive sign when this quantity
is squared. Forgetting this fact would yield a value for v that is smaller than vx, but the
magnitude of a vector cannot be smaller than either of its components. 5. SSM REASONING AND SOLUTION x = r cos θ = (162 km) cos 62.3° =
y = r sin θ = (162 km) sin 62.3° = 75.3 km
143 km ____________________________________________________________________________________________ 6. REASONING AND SOLUTION The horizontal displacement is
x = 19 600 m – 11 200 m = 8400 m The vertical displacement is Chapter 3 Problems 101 y = 4900 m – 3200 m = 1700 m
The magnitude of the displacement is therefore,
2
2
∆r = x 2 + y 2 = ( 8400 m ) + (1700 m ) = 8600 m ____________________________________________________________________________________________ 7. SSM REASONING The displacement of the elephant seal
has two components; 460 m due east and 750 m downward.
These components are mutually perpendicular; hence, the
Pythagorean theorem can be used to determine their resultant. 460 m R 750 m SOLUTION From the Pythagorean theorem,
R 2 = (460 m) 2 + (750 m) 2
Therefore,
R = ( 460 m) 2 + ( 750 m) 2 = 8.8 × 10 2 m 8. REASONING Consider first the shopper’s ride up the escalator. Let the diagonal length of
the escalator be L, the height of the upper floor be H, and the angle that the escalator makes
with respect to the horizontal be θ (see the diagram). Because L is the hypotenuse of the
right triangle and H is opposite the angle θ, the
three quantities are related by the inverse sine
s
function:
L
H ho H
−1
−1 θ = sin = sin (1.4)
θ
L
D
L
h
Up the escalator Now consider the entire trip from the bottom to
the top of the escalator (a distance L), and then
Entire view
from the top of the escalator to the store
entrance (a distance s). The right turn between
these two parts of the trip means that they are perpendicular (see the diagram). The
shopper’s total displacement has a magnitude D, and this serves as the hypotenuse of a right
triangle with L and s. From the Pythagorean theorem, the three sides are related as follows:
D 2 = L2 + s 2 . SOLUTION Solving D 2 = L2 + s 2 for the length L of the escalator gives L = D 2 − s 2 . We
H
now use this result and the relation θ = sin −1 to obtain the angle θ:
L 102 KINEMATICS IN TWO DIMENSIONS θ = sin 9. −1 H L H −1 = sin 2
2 D −s −1 = sin 6.00 m (16.0 m )2 − ( 9.00 m )2 = 27.0o SSM WWW REASONING
a. We designate the direction down and parallel to the ramp as the +x direction, and the
table shows the variables that are known. Since three of the five kinematic variables have
values, one of the equations of kinematics can be employed to find the acceleration ax. xDirection Data
x ax vx v0x +12.0 m ? +7.70 m/s 0 m/s b. The acceleration vector points down and parallel to the ramp, and the
angle of the ramp is 25.0° relative to the ground (see the drawing).
Therefore, trigonometry can be used to determine the component aparallel
of the acceleration that is parallel to the ground. t ax
25.0° aparallel SOLUTION
2
2
a. Equation 3.6a ( vx = v0 x + 2ax x ) can be used to find the acceleration in terms of the three
known variables. Solving this equation for ax gives
ax = 2
2
vx − v0 x 2x ( +7.70 m/s ) − ( 0 m/s )
=
2 ( +12.0 m )
2 2 = 2.47 m/s 2 b. The drawing shows that the acceleration vector is oriented 25.0° relative to the ground.
The component aparallel of the acceleration that is parallel to the ground is aparallel = ax cos 25.0° = ( 2.47 m/s 2 ) cos 25.0° = 2.24 m/s 2 10. REASONING AND SOLUTION
a. Average speed is defined as the total distance d covered divided by the time ∆t required
to cover the distance. The total distance covered by the earth is onefourth the circumference
of its circular orbit around the sun:
1 d = 4 x 2π (1.50 x 1011 m) = 2.36 x 1011 m v= d 2 .36 × 1011 m
=
= 2.99 × 10 4 m / s
6
∆t
7.89 × 10 s Chapter 3 Problems 103 b. The average velocity is defined as the displacement
divided by the elapsed time.
In moving onefourth of the distance around the sun,
the earth completes the displacement shown in the
figure at the right. From the Pythagorean theorem,
the magnitude of this displacement is ∆r
r
r ∆r = r2 + r2 = 2r Thus, the magnitude of the average velocity is
v= ∆r
=
∆t 2 × 1.50 × 10 11 m
= 2.69 × 10 4 m / s
7.89 × 10 6 s The component method can be used to determi...
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 Spring '13
 CHASTAIN
 Physics, The Lottery

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