Train starts from rest, hence the initial velocity u = 0.
It moves with acceleration = 2m/s2 for half minute (30 seconds).
Distance covered in this time interval is given by:
Velocity attained by this acceleration after 30 seconds:
From this velocity, brakes are applied and train comes to rest in 60 seconds.
The retardation is given by:
Distance covered in this time:
sv<sup>2</sup>= u<sup>2</sup> + 2as
So, total distance moved =900m+1800m=2700m.
Maximum speed of the train=60m/s.
Position of the train at half its maximum speed.
Here, you need to note that first the train is accelerating to 60 m/s, and then it is decelerating to 0 m/s. So there are two positions when speed is 30 m/s.
i. When the train is accelerating with an acceleration of 2 m/s,
time at which speed = 30m/s is:
At 15s, distance covered from origin is:
ii. When the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s<sup>2</sup> , time at which the speed reaches 30 m/s is:
At 30s, distance covered is:
=1350m (from the initial 900m covered).
So, distance from origin =900+1350m=2250m.