A car is parked on a cliff overlooking the ocean on an incline that makes an ang
ID: 1693251 • Letter: A
Question
A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 18.0° below the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline with a constant acceleration of 2.90 m/s2 for a distance of 35.0 m to the edge of the cliff, which is 50.0 m above the ocean.(a) Find the car's position relative to the base of the cliff when the car lands in the ocean.
m
(b) Find the length of time the car is in the air.
s
Explanation / Answer
Distance covered on the cliff S = 35 m Angle of inclination ? = 18 degrees Acceleration a = 2.9 m/s^2 Time taken to leave the edge of the cliff (t') : From kinematic relation S = (1/2)at'^2 35 = (0.5)(2.9)t'^2 t'^2 = 24.13 t' = 4.91 s Velocity of the car when leaves the incline v = at = 14.24 m/s The verticle component of v is, v sin18 = 4.4 m/s The horizontal component of v is, v cos18 = 13.54 m/s From kinematic relation we caqn write H(t) = (v sin18)t + (1/2)gt^2 50 = 4.4t + 4.9t^2 4.9t^2 + 4.4 t - 50 = 0 On solving the above quadratic equation (b) time t = 2.77 s (a) Car position with respect to the base of the cliff = v cos18 * t = 13.54*2.77 = 37.5 m (a) Car position with respect to the base of the cliff = v cos18 * t = 13.54*2.77 = 37.5 m (a)37.5 m , (b)2.77 sRelated Questions
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