A certain cable car in San Francisco can stop in 10 seconds when traveling at ma

Question

A certain cable car in San Francisco can stop in 10 seconds when traveling at maximum speed. On one occasion, the driver sees a dog a distance x meters in front of the car and slams on the brakes instantly. The car reaches the dog 8 seconds later, and the dog jumps off the track just in time. If the car travels 4.0 meters beyond the position of the dog before coming to a stop, what can you find out about the cable car and the dog in terms of the initial velocity of the car, the postition of the dog, the acceleration of the car and the time.

Explanation / Answer

Let the initial maximum velocity of the car is u m/s,
=>By v = u - at
=>0 = u - a x 10
=>u = 10a ---------(i)
The car comes to the halt after (d+4) meter distance and take 8 sec to travel d meter,
=>By s = ut - 1/2at^2
=>d = 10a x 8 - 1/2 x a x (8)^2
=>d = 80a - 32a
=>d = 48a -----------(iii)
By v^2 = u^2 - 2as
=>0 = (10a)^2 - 2 x a x (d+4)
=>100a^2 = 2ad + 8a
=>100a^2 = 2a x 48a + 8a
=>4a^2 - 8a = 0
=>4a(a -2) = 0
=>a = 0 or 2 m/s^2
=>as a 0
=>a = 2 m/s^2
By putting this value in (ii) :-
=>d = 48 x 2 = 96 m

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