$\begin{array}{1 1}(A)\;\large\frac{S}{a} \\(B)\;\large\frac{2S}{a}\\(C)\;2\sqrt{\large\frac{S}{a}}\\(D)\;\sqrt {\large\frac{S}{2a}} \end{array} $

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If the train has accelerated with acceleration $\alpha$ for t , seconds , moved with zero acceleration $\alpha$ (ie) constant velocity nv for t_o seconds and decelerated with retardation $\beta$ for $t_2$ seconds .

$t= t_1+t_0+t_2$

Where l is the distance covered with constant velocity v.

The above relation shows that for minimum time, $ \alpha \to a, \beta \to \alpha $ and $l \to 0 . Thus the minimum time is given by

$t_{min} = \large\frac{v}{a} $$+0+\frac{v}{a}$$=\large\frac{2v}{a}$

also $S= \large\frac{1}{2} $$(t_{min}) C$

Eliminating v from (i) and (ii)

$2\sqrt{\large\frac{S}{a}}$

Hence C is the correct answer.

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