For a train engine moving with speed of 20 ms ^ – 1 , the driver...

Given that, the initial velocity $u = 20m / s$, the distance of the station ${S}_{1} = 500m$, 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

${v}^{2} = {u}^{2} - 2aS . . . . . . . . . . . . . . . . . . . . . . . . ( 1 )$

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.

$\begin{cases}0 & = & ( 20{)}^{2} - 2 \times a \times 500 \\ a & = & 0 . 4m / {s}^{2}\end{cases}$

For the second scenario, the initial velocity $u = 20m / s$ and the distance travelled by the train ${S}_{2} = 250m$.

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

$\begin{cases}{v}^{2} & = & ( 20{)}^{2} - 2 \times 0 . 4 \times 250 \\ & = & 200 \\ v & = & \sqrt{200} . . . . . . . . . . . . . . . . . . . . . . . . . . ( 2 )\end{cases}$

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