Question

The greatest acceleration or deceleration that a train may have is ${}^{\prime }{a}^{\prime }$. The minimum time in which the train can get from one station to the next at a distance ${}^{\prime }{s}^{\prime }$ is –

• A. $\sqrt{\frac{s}{a}}$
• B. $\sqrt{\frac{2s}{a}}$
• C. $\frac{1}{2}\sqrt{\frac{s}{a}}$
• D. $2\sqrt{\frac{s}{a}}$

Answer 1

The correct option is D $2\sqrt{\frac{s}{a}}$

Here, we use the symmetry of motion i.e., from ending point if you go reverse in time, it looks like train is accelerating. So, here for minimum time, train should accelerate for half the distance and then decelerate.
For minimum time, the train must accelerate fast as well as decelerate fast without moving with constant velocity.
As acceleration and deceleration are of equal magnitude so, train must accelerate as well as retard for same distance.

$v=u+a{t}_1$
$v_{max}=a{t}_1$
$0=v_{max}-a{t}_2$
$=a{t}_1=a{t}_2$
$t_1=t_2$
$\frac{s}{2}=\frac{1}{2}a{t}_1^2$
$t_1=\sqrt{\frac{s}{a}}$
$T=2t_1$
$\Rightarrow T=2\sqrt{\frac{s}{a}}$