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A cyclist driving at 36 km/h stops in 2s by the application of brakes.Calculate : (1) retardation (2) distance covered during application of brakes ?
Asked by Vikas | 23 May, 2018, 12:20: AM
Equations are to be applied is "v = u - a×t " and "S = u×t - (1/2)×a×t2 "
v is final velocity, u initial velocity, a is retardation, S is distance travelled and t is time.

initial velocity = 36 km/h ×(5/18) = 10 m/s

(1) retardation :-  0 = 10-a×2 , solving for a  we get  retardation a = 5 m/s2

(2) distance :- S = 10×2 -(1/2)×5×4 = 10 m
Answered by Thiyagarajan K | 23 May, 2018, 11:14: AM

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