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a car moving at certain speed stops on applying brakes with in 16 m.if the speed of the car is doubled,maintaining the same retardation.then at what distance does it stop?Also calculate the percentage change in this distance.
Asked by bairava6324 | 19 Dec, 2019, 05:52: PM

### Let u m/s be the initial speed. To stop the car in 16 m, required retardation a is given bya = u2 / (2×16) = u2 /32 ..............................(1)After doubling the speed and maintaining the same retardation, distance S travelled before stopping the car  is given byS = (2u)2 /(2×a) =  2×u2 / a ..............(2)substituting for retardation a from (1) in (2),  S = 2×u2 / (u2 / 32) = 64 mThe percentage change in this distance is =[(v2 - v1)/ v1] x 100=  [(64 - 16)/ 16] x 100 = 3 x 100 = 300 %

Answered by Shiwani Sawant | 19 Dec, 2019, 06:22: PM

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