A train starts fr

a. Here initial velocity, u=0

Let v be the final velocity

Acceleration, a=2 m/s²

Now, at t=30 s

We know,

When breaks are applied u1=60 m/s

v1=0

t=60 s

Total distance moved is

b. Maximum speed attained by the train is v= 60 m/s

c. Half the maximum speed=60/2=30 m/s

When acceleration of the train, a=2 m/s²

When the train is decelerating with, a1=-2 m/s²

Position from the starting point is 900+1350=2250 m=2.25 km