A train starts from a station and stops at another station. First the velocity of the train increases at a uniform rate, reaches a maximum velocity v and then starts decreasing at a constant rate. If the time spent for the entire journey is t, prove that the distance between the two stations is `(1)/(2)`vt.

A train begins its journey from a station with a uniform acceleration of 0.2 "m.s"^(-2) and moves for 1 min. Next, it moves for a further 5 min at uniform velocity. It then appIies brakes producing a retardation of 0.3 "m.s"^(-2) and stops at the next station. What is the distance between the two station?

Starting from rest a body moves at first with constant acceleration a. Then it moves with a uniform velocity and finally comes to rest with a constant retardation a. If the displacement is s and time taken is t, prove that the body was in motion with uniform velocity for a time interval of sqrt(t^(2)-4s//a).

A train begins its journey from station A and stops at station B after 45 min. C is a certain point between A and B where the train attains its maximum velocity of 50 km. "h"^(-1) . If the train travels from A to C with a uniform acceleration and from C to B with a uniform retardation, calculate the distance between A and B.

The distance between two station is d km. A train went the second station from the first station with uniform velocity. If the velocity of the train would be a km more, then the time taken by the train would be t hours lesser than before.

A train moves with uniform velocity u while going from station p to station Q. It returns along the same path with uniform velocity V. Its average velocity during the journey is :

A train travels from station A to station B at a constant speed of 40 km. h^(-1) and returns from B to A at 60 km . h^(-1) . Find the average speed and average velocity of the train.