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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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The greatest acceleration or deceleration that a train may have is $\alpha$ .The minimum time in which the train can get from one station to the next at a distance S is

$\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.
answered Jul 10, 2014 by meena.p
 

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