Answer to Question #287952 in Physics for jake

Question

A KTM commuter train travels between two stations, A and B. The train starts from rest from station A and accelerates at a constant rate for 0.5 km until it reaches a speed of 30 m/s. It then travels at this constant speed for T s before decelerating at a constant rate of 1.5 m/s² and coming to rest at station B. Figure 1 shows the velocity-time graph of the train’s journey between the two stations. Determine the value of T if the distance between the two stations 6 km. Calculate also the total time taken for the whole journey.

Answer 1

The distance covered during acceleration: 500m.

The distance covered during constant speed motion: 30T.

The distance covered during deceleration:

D=\dfrac{30^2}{2•1.5}=300 m.

The entire trip is (800+30T) m.