A certain cable car in San Francisco can stop in 10 s when traveling at maximum speed.?
A certain cable car in San Francisco can stop in 10 s when traveling at maximum speed. On one occasion, the driver sees a dog a distance d m in front of the car and slams on the brakes instantly. The car reaches the dog 7.44 s later, and the dog jumps off the track just in time. If the car travels 3.67 m beyond the position of the dog before coming to a stop, how far was the car from the dog? [Hint: You will need three equations.]
Fireman2013-01-12T15:36:19Z
Favorite Answer
Let the initial velocity of the car is u m/s =>By v = u - at =>0 = u - 10a =>u = 10a ---------------(i) By s = ut - 1/2at^2 =>d = 10a x 7.44 - 1/2 x a x (7.44)^2 =>d = 46.72a ------------(ii) By v^2 = u^2 - 2as =>0 = (10a)^2 - 2 x a x (46.72a + 3.67) =>100a^2 - 93.44a^2 - 7.34a = 0 =>a(6.56a - 7.34) = 0 =>a = 1.12 m/s^2 Thus d = 46.72 x 1.12 = 52.28 m
RE: A certain cable car in San Francisco can stop in 10 s when traveling at maximum speed.? A certain cable car in San Francisco can stop in 10 s when traveling at maximum speed. On one occasion, the driver sees a dog a distance d m in front of the car and slams on the brakes instantly. The car reaches the dog 7.44 s later, and the dog jumps off the track just in time. If the car travels...
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