The engine of a train can impart a maximum acceleration of $1 m s^{- 2}$ and the brakes can give a maximum retardation of $3 m s^{- 2} .$ The least time during which a train can go from one place to the other place at a distance of 1.2 km is nearly

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108 s

191 s

56.6 s

Time is fixed

answer is C.

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Detailed Solution

As, $1 = \frac{v}{t_{1}} a n d 3 = \frac{v}{t_{2}}$

$∴ 1200 = \frac{1}{2} (t_{1} + t_{2}) v,$

$1200 = \frac{1}{2} (v + \frac{v}{3}) v = \frac{1}{2} \frac{4 v^{2}}{3} = \frac{2 v^{2}}{3}$

$v^{2} = 1800$

$1200 = \frac{1}{2} t \times \sqrt{1800}$

$t = \frac{2400}{\sqrt{1800}} s$

$= \frac{2400}{42.43} s = 56.6 s$

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