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 √x ms⁻¹. The value of x is ____________.
(Assuming same retardation is produced by brakes)

Solution

Physics sir · 3 steps

Get the retardation from the full 500 m stop, then use it again over the shorter 250 m.

1. Find the retardation. With v² = u² - 2as, the train goes from 20 m s⁻¹ to rest in 500 m. So 0 = 400 - 2a(500), which gives a = 0.4 m s⁻².
2. Shorter braking distance. Now the brakes act over only 250 m with the same a. The train cannot lose all its speed, so it reaches the station still moving.

   v² = u² - 2as = 400 - 2 × 0.4 × 250 = 400 - 200 = 200

3. So. The speed at the station is v = √200 m s⁻¹, so x = 200.

The common slip: 14. Reporting the speed √200 ≈ 14.1 m s⁻¹ gives 14. The question asks for x, which is v² = 200.

The idea to keep. With constant retardation, the speed lost depends on braking distance through v², not v. Halve the distance and you remove half of u², not half of u.

Answer from the NTA final answer key. Solution by PrepWiser.