1. Given data

• Initial speed of train engine: $u = 20\,\text{m s}^{-1}$
• To stop at the station, brakes must be applied $500\,\text{m}$ before the station.
• Retardation produced by brakes is the same in both cases.

2. Find the retardation using the first situation

When the train comes to rest exactly at the station:

$v = 0, \quad u = 20\,\text{m s}^{-1}, \quad s = 500\,\text{m}$

Using the equation of motion,

$v^2 = u^2 + 2as$

Substitute the values:

$0 = (20)^2 + 2a(500)$

$0 = 400 + 1000a$

$a = -\frac{400}{1000} = -0.4\,\text{m s}^{-2}$

So, the retardation is

$0.4\,\text{m s}^{-2}$

3. Second situation: brakes applied at half the distance

Half of $500\,\text{m}$ is

$250\,\text{m}$

Now the engine starts braking $250\,\text{m}$ before the station, with the same initial speed $u=20\,\text{m s}^{-1}$ and same acceleration $a=-0.4\,\text{m s}^{-2}$.

Let the speed at the station be $v$.

Again use:

$v^2 = u^2 + 2as$

$v^2 = (20)^2 + 2(-0.4)(250)$

$v^2 = 400 - 200 = 200$

Given that the speed at the station is $\sqrt{x}\,\text{m s}^{-1}$, so

$v = \sqrt{x}$

Thus,

$x = v^2 = 200$

4. Final answer

$\boxed{200}$

5. Comparison with stored correct answer

Stored correct answer = $200$

Our derived answer also = $200$.

So, the answer agrees with the stored correct answer.