Question

A train start from rest accelerates uniformly so that it attain a speed of 400-1 in 20s. It Travel at this speed for 40s and is then brought to rest with uniform retardation after 30s

What is the

The total distance covered

The retardation of the train

Answer 1

The total distance traveled by a train

$$d=d_1+d_2+d_3$$

$$d=\frac{v+0}{2}t_1+vt_2+\frac{0+v}{2}t_3$$

$$d=\frac{400}{2}\times20+400\times 40+\frac{400}{2}\times 30\\ =26000\:\rm m$$

The retardation of the train

$$a_3=\frac{\Delta v_3}{\Delta t_3}=\frac{-400}{30}=-13.3\;\rm m/s^2$$