3. A car is parked on a steep incline, making an angle of 35.0° below the horizo

Question

3. A car is parked on a steep incline, making an angle of 35.0° below the horizontal and overlooking the ocean, when its brakes fail and it begins to roll. Starting from rest at t-0, the car rolls down the incline with a constant gravitational acceleration, traveling 50.0 m to the edge of a vertical cliff. The cliff is 30.0 m above the ocean. a. What is the speed of the car when it reaches the edge of the cliff? b. How much time does it take for the car to reach the edge of the cliff? c. What are the horizontal and vertical components of the car's velocity when it lands in the ocean? d. How much time elapses between the moment the car begins to roll to when it lands in the ocean? e. How far from the base of the cliff does the car land in the ocean? [Answers: (a) 23.7 m/s along the incline (b) 4.22 s (c) (19.4,-278) m/s (d) 5.67 s (e) 28.1 m]

Explanation / Answer

3] Acceleration of the car is:

a = gsin35 = 5.62 m/s2 downwards

initial velocity = u = 0 m/s

use v2 = u2 + 2aS

v = [2 x 5.62 x 50]1/2 = 23.71 m/s

b] S = (1/2)at2

50 = (1/2)5.62t2

=> t = 4.22 s

this is the time it takes to reach the edge of the cliff.

c] ux = 23.71cos35 = 19.42 m/s

uy = 13.6 m/s

and a = g = 9.8 m/s2

vy = [13.62 + 2(9.8)(30)]1/2 = 27.80 m/s

d] vy = uy + gt

27.8 = 13.6 + 9.8t

=> t = 1.45s

so, the time taken for the car to land in the ocean from the time it begins to roll is:

T = 4.22 + 1.45 = 5.67 s

e] Lx = uxt = 19.4 x 1.45 = 28.13 m.

this is the distance the car travels horizontally.

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