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CBSE Class 10 Answered

A car is moving at a certain speed stops on applying within 16 m. If the speed of car is doubled, maintaining the same retardation, then at what distance does it stops? Also calculate the percentage change in distance.
Asked by acv27joy | 12 May, 2018, 08:26: PM
answered-by-expert 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
Answered by Thiyagarajan K | 13 May, 2018, 12:17: PM

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