Answer to Question #188138 in Mechanical Engineering for Vikas Yadav

Question

Two stations A and B are 10 km apart in a straight track, and a train starts from A and comes to rest at B. For three quarters of the distance, the train is uniformly accelerated and for the remainder distance uniformly retarded. If it takes 15 minutes over the whole journey, find its acceleration, retardation and the maximum speed it attains. 

Answer 1

The maximum speed it attains.

$S=\frac{D}{T}=\frac{10*1000}{15*60}=11.11 m/s$

Acceleration

$t= \frac{3/4*10}{11.11}=0.675 s$

$s=ut +\frac{1}{2}at^2$

$\frac{3}{4} *10=\frac{1}{2}a*0.675^2 \implies a= 32.92 m/s^2$

Retardation

 $\frac{1}{4} *10=\frac{1}{2}a *899.325^2 \implies a= 0.00000618211 m/s^2$