A car is parked on a steep incline, making an angle of 37.0° below the horizonta

Question

A car is parked on a steep incline, making an angle of 37.0° 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 acceleration of 7.9 m/s2, traveling 63.2 m to the edge of a vertical cliff. The cliff is 25.9 m above the ocean. Find

(a) the speed of the car when it reaches the edge of the cliff.  m/s

(b) the time interval elapsed when it arrives there.  s

(c) the velocity of the car when it lands in the ocean.  i +  j m/s

(d) the total time interval the car is in motion.  s

(e) the position of the car when it lands in the ocean, relative to the base of the cliff.  m

Explanation / Answer

A) By third equation of motion,

v = sqrt (2as) = sqrt (2*7.9*63.2)

= 31.6 m/s

B) time = v/a = 31.6/7.9

= 4 s

C) Vx = 31.6*4/5 =25.28 m/s

Vy = - sqrt((31.6*3/5)^2 + 2*9.8*25.9)

= - 29.45 m/s

V = 25.28 i - 29.45 m/s

D) t = 4 + (29.45- 31.6*3/5)/9.8

= 4 + 1.07

= 5.07 s

E) d = Vx*(5.07-4) = 25.28*1.07

= 27.05 m

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