To solve this problem, we can break it down into two parts: the acceleration phase and the deceleration phase.
Given: Distance between subway stops = 1100 m Acceleration during the first half = 1.2 m/s^2 Deceleration during the second half = -1.2 m/s^2
(a) Travel time: Let's calculate the time it takes to cover the first half of the distance using the formula: distance = (initial velocity * time) + (0.5 * acceleration * time^2)
Since the train starts from rest, the initial velocity is 0 m/s.
For the first half of the distance: distance = 550 m (half of 1100 m) acceleration = 1.2 m/s^2
550 = (0 * t) + (0.5 * 1.2 * t^2) 550 = 0 + 0.6t^2 0.6t^2 = 550 t^2 = 550 / 0.6 t^2 = 916.67 t ≈ √916.67 t ≈ 30.28 seconds (approx.)
The time it takes to cover the second half of the distance will be the same since the deceleration is symmetrical to the acceleration. Therefore, the total travel time is: Total travel time = 2 * 30.28 seconds Total travel time ≈ 60.56 seconds (approx.)
(b) Maximum speed: To find the maximum speed, we need to determine the velocity at the end of the acceleration phase, which will be the same as the initial velocity for the deceleration phase.
Using the formula: final velocity = initial velocity + (acceleration * time)
For the first half of the distance: initial velocity = 0 m/s (train starts from rest) acceleration = 1.2 m/s^2 time = 30.28 seconds (from part a)
final velocity = 0 + (1.2 * 30.28) final velocity ≈ 36.34 m/s (approx.)
The maximum speed of the train is approximately 36.34 m/s.