A train is travelling with a velocity of $72 {km h}^{- 1} .$ The brakes are applied to retard the motion of the train uniformly. If the train is stopped 50 meter away from the place where the brakes were applied, then the retardation of the train is

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$- 3 {ms}^{- 2}$

$- 4 {ms}^{2}$

$2 {ms}^{- 2}$

$6 {ms}^{- 2}$

answer is B.

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Detailed Solution

$u = 72 km/h,s = 50 m, v = 0$

$u = \frac{72 \times 5}{18} = 20 m/s$

$v^{2} - u^{2} = 2as$

$a = \frac{v^{2} - u^{2}}{2 a} = \frac{0 - 400}{2 \times 50} = \frac{- 400}{100}$

$= - 4 {m/s}^{2}$

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