Starting from rest, a train travels with a constant acceleration of 2.0 m/s ^2 for half a minute. Subsequently, the brakes are applied, bringing the train to rest in one minute. Determine the position(s) of the train at half the maximum speed.

Step-by-step solution

Maximum speed attained by the train is v_ max = u + a₁ t₁ = 0 + 2.0 30 = 60 m/s . Half of the maximum speed is v = 60 2 = 30 m/s . The train attains this speed twice: once during acceleration and once during deceleration. During the acceleration phase, the position s₁ at v = 30 m/s is given by: v^2 = u^2 + 2a₁ s₁ 30^2 = 0 + 2(2.0)s₁ s₁ = 900 4 = 225 m The total distance covered during the acceleration phase is: S₁ = 1 2 a₁ t₁^2 = 1 2 (2.0)(30)^2 = 900 m During the deceleration phase, the train comes to rest from 60