For a train engine moving with speed of 20   ms – 1 , the driver must apply brakes at a distance…

Given that, the initial velocity $u=20 \mathrm{m}/\mathrm{s}$, the distance of the station $S_1=500 \mathrm{m}$, and the final velocity $v=0$.

The relation between the distance travelled $\left(S\right)$ by the train with its initial velocity $\left(u\right)$ and its final velocity $\left(v\right)$ is given by

$v^2= u^2-2aS........................\left(1\right)$

where, $a$ indicates the retardation of the train.

Substitute the values corresponding to the initial condition into equation (1) and solve to calculate the retardation of the train.

$\begin{array}{rcl}0& =& (20{)}^2-2\times a\times 500\\ a& =& 0.4 \mathrm{m}/{\mathrm{s}}^2\end{array}$

For the second scenario, the initial velocity $u=20 \mathrm{m}/\mathrm{s}$ and the distance travelled by the train $S_2=250 \mathrm{m}$.

Use these values along with the value of the retardation into equation (1) and solve to calculate the final velocity of the train.

$\begin{array}{rcl}{v}^2& =& (20{)}^2-2\times 0.4\times 250\\ & =& 200\\ v& =& \sqrt{200}..........................\left(2\right)\end{array}$

Comparing the value of the final velocity obtained in equation (2) with the given expression, we get $x= 200$.