The greatest acceleration or deceleration that a train may have is $a$ . The minimum time in which the train can go from one station to the next at a distance $S$ is

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

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\sqrt{\frac{S}{a}}

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

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\sqrt{\frac{a}{S}}

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Solution

The correct option is A $2 \sqrt{\frac{S}{a}}$
There is no limit on the speed of the train, so minimum time will be when the train will accelerate for $\frac{T}{2}$ and deaccelerate for $\frac{T}{2}$ .
So time taken will calculated by $S = u t + \frac{1}{2} a t^{2}$ , where $u = 0$ , $d i s p l a c e m e n t = \frac{S}{2}, t i m e = \frac{T}{2}$ , same distance will be covered while acceleration as well as deacceleration.
So $T = 2 \sqrt{\frac{S}{a}}$

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The greatest acceleration or deceleration that a train may have is a. The minimum time in which the train can go from one station to the next at a distance S is

The greatest acceleration or deceleration that a train may have is $a$ . The minimum time in which the train can go from one station to the next at a distance $S$ is