For a train engine moving with speed of $$20$$ 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 $$\sqrt{x}$$ ms$$^{-1}$$. The value of $$x$$ is ______. (Assuming same retardation is produced by brakes)

Scenario 1: Full stopping distance

Initial velocity $$u = 20 \text{ m/s}$$

Final velocity $$v_1 = 0 \text{ m/s}$$

Displacement $$s_1 = 500 \text{ m}$$

$$v_1^2 - u^2 = 2as_1$$

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

$$-400 = 1000a$$

$$a = -0.4 \text{ m/s}^2$$

Scenario 2: Half the stopping distance

The brakes are now applied at half the distance, meaning the train travels $$s_2 = 250 \text{ m}$$ while decelerating. We want to find its speed $$v_2$$ as it crosses the station.

Initial velocity $$u = 20 \text{ m/s}$$

Acceleration $$a = -0.4 \text{ m/s}^2$$

Displacement $$s_2 = 250 \text{ m}$$

$$v_2^2 - u^2 = 2as_2$$

$$v_2^2 - (20)^2 = 2(-0.4)(250)$$

$$v_2^2 - 400 = -0.8(250)$$

$$v_2^2 - 400 = -200$$

$$v_2^2 = 200$$

$$v_2 = \sqrt{200} \text{ m/s}$$