The train starts to brake at a speed of 20m / s. Determine its speed after passing 2/3 of the braking distance.

Given: V (speed of a given train at the beginning of braking) = 20 m / s; S1 (distance traveled) = 2/3 S (braking distance).

Since the train passed 2/3 of its stopping distance, its kinetic energy decreased by 2/3: A1 = F * S1 = F * 2/3 S = 2/3 A and Ek1 = Ek – ΔEk = A – A1 = A – 2/3 A = 1/3 A.

Therefore, Ek1 = m * V1 ^ 2/2 = 1/3 * m * V2 / 2, whence V1 = √ (1/3 * V2) = √ (1/3 * 20 ^ 2) = 11.547 m / s ≈ 12 m / s.

Answer: After passing 2/3 of the braking distance, the given train will have a speed of 12 m / s.