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:

S=ut+1/2 at<sup>2</sup>

=0+1/2×2×30×30

=900m

Velocity attained by this acceleration after 30 seconds:

v=u+at

v=0+2×30

v=60m/s

From this velocity, brakes are applied and train comes to rest in 60 seconds.

The retardation is given by:

v=u–at

0=60–a×60

a=1m/s<sup>2</sup>

Distance covered in this time:

sv<sup>2</sup>= u<sup>2</sup> + 2as

0=(60)2+2(−1)S

0=3600–2S

s=3600/2=1800m.

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:

v=u+at

30=0+2xt

t=15s

At 15s, distance covered from origin is:

S=ut+1/2at<sup>2</sup>

=0+1/2×2×15×15

=225m

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:

v=u–at

30=60–1xt

t=30s

At 30s, distance covered is:

S=ut–1/2at<sup>2</sup>

=60×30-1/2x1x(30)2

=1800–(15×30)

=1800–450

=1350m (from the initial 900m covered).

So, distance from origin =900+1350m=2250m.