A car moving with a velocity of 20 m s^-1 is stopped in a distance of 40 m. If the same car is travelling at double the velocity, the distance travelled by it for same retardation is

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Correct answer: D. (d) : Initial velocity, u=20 m s^-1; Final velocity, v=0 m s^-1; Distance travelled, s=40 m Using third equation of motion, v^2-u^2=2 as ;0-20^2=2 a 40 We get, a=-5 m s^-2 For second case, u=2 20 m s^-1=40 m s^-1, v=0 m s^-1 and a=-5 m s^-2 Using the third equation of motion, s= 0- (40 )^22 (-5 )=160 m ∴ Distance travelled by the same car travelling at double velocity is 160 m.

$A car moving with a velocity of 20 m s^{- 1} is stopped in a distance of 40 m . If the same car is travelling at double the velocity, the distance travelled by it for same retardation is$

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