Question

The maximum acceleration or deceleration that a train may have is $a=5m{s}^{-2}$. The minimum time in which the train may reach from one station to the other separated by a distance $d=500m$ is $t_0=5ns$. The value of $n$ is______.

Answer 1

Answer: 4

From the graph,

$d=\frac{1}{2}\times v_0{t}_0=\frac{v_0{t}_0}{2}$
and $\tan \theta =a=\frac{v_0}{t_0/2}=\frac{2v_0}{t_0}$
$\therefore v_0=\frac{a{t}_0}{2}$
$\therefore d=\frac{t_0}{2}\times \frac{a{t}_0}{2}=\frac{a{t}_0^2}{4}$
$\therefore t_0=2\sqrt{\frac{d}{a}}=2\sqrt{\frac{500}{5}}$
$=20s=20=5n$
$\therefore n=4$