CBSE Class 9 Answered
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
Expert Answer
Let u m/s be the initial speed. To stop the car in 16 m, required retardation a is given by
a = u2 / (2×16) = u2 /32 ..............................(1)
After doubling the speed and maintaining the same retardation, distance S travelled before stopping the car is given by
S = (2u)2 /(2×a) = 2×u2 / a ..............(2)
substituting for retardation a from (1) in (2), S = 2×u2 / (u2 / 32) = 64 m
The 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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