For a train engine moving with speed of $$20 \text{ ms}^{-1}$$, the driver must apply brakes at a distance of 500 m before the station for the train to come to rest at the station. If the brakes were applied at half of this distance, the train engine would cross the station with speed $$\sqrt{x} \text{ ms}^{-1}$$. The value of $$x$$ is ______. (Assuming same retardation is produced by brakes)

We need to determine the value of $$x$$ such that the speed of the train as it crosses the station is $$\sqrt{x}$$ m/s.

Since the train comes to rest after travelling 500 m from its initial speed, using the kinematic relation $$v^2 = u^2 - 2as$$ gives $$0 = 20^2 - 2a(500)$$, which yields $$a = 0.4$$ m/s².

Brakes are applied at half the distance to the station, namely 250 m, and the train decelerates over this stretch. Substituting into $$v^2 = 20^2 - 2(0.4)(250)$$ gives $$v^2 = 400 - 200 = 200$$.

From this result, the speed at the station is $$\sqrt{200}$$ m/s, so that $$x = 200$$. Therefore, the answer is 200.