The greatest acceleration or deceleration that a train may have is a. The minimum time in which the train may reach from one station to the other separated by a distance is-
A. $\sqrt {\dfrac{d}{a}} $
B. $\sqrt {\dfrac{{2d}}{a}} $
C. $\dfrac{1}{2}\sqrt {\dfrac{d}{a}} $
D. $2\sqrt {\dfrac{d}{a}} $

Hint: In order to find this numerical, we need to draw the velocity- time graph in which the train which accelerates and decelerates. After taking the formula of the area of the triangle we can calculate the distance.

Complete step by step answer:
Distance is equal to the Area under the velocity- time graph on the axis of time.
Let us consider the total distance travelled by train is taken as d. It covers the first half with an acceleration and the second half it decelerates therefore it takes half time for both the cases.

Train is initially at rest therefore the initial velocity u = 0.
By using Area of triangle,
Distance=$\dfrac{1}{2} \times base \times height$
$
  d = \dfrac{1}{2} \times t \times \dfrac{1}{2}at \\
\Rightarrow d = \dfrac{1}{4}a{t^2} \\
  \Rightarrow {t^2} = \dfrac{{4d}}{a} \\
  \Rightarrow t = \sqrt {\dfrac{{4d}}{a}} \\
  \Rightarrow t = 2\sqrt {\dfrac{d}{a}} \\
 $

Hence, the correct option is D.

Note:
To solve the numerical based on equation of motion we should remember the few points:
> If the body is at rest then we should take the initial velocity as zero.
> If the body stops after sometime then final velocity is taken as zero.
> If the body moves with uniform velocity then acceleration is taken as zero.