max speed = 2*30 = 60m/s
let retardation is a
then
60 – a*60 = 0
so a = 1m/s
time when train at half the max speed
30 = 2*t
t = 15 sec
position at time t = 15
= ½*2*152 = 225 m
total distance covered by train
d = ( ½*2*302 ) + (60*60 – ½*1*602)
d = (900) + (3600 – 1800) = 2700m
Hope it clears.
						

							
HELLO!
 
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 + ½ at2
= 0 + ½ x 2 x 30 x 30
= 900m
 
Velocity attained by this acceleration after 30 seconds:
v = u + at
=> v = 0 + 2 x 30
=> v = 60 m/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 x 60
=> a = 1 m/s2
 
Distance covered in this time:
v2 = u2 + 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 = 60 m/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 + 2 x t
=> t = 15s
 
At 15s, distance covered from origin is:
S = ut + ½ at2
= 0 + ½ x 2 x 15 x 15
= 225m
 
(II) When the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 ,  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 – ½ at2
= 60 x 30 – ½ x 1 x (30)2
= 1800 – (15 x 30)
= 1800 – 450
= 1350m (from the initial 900m covered).
 
So, distance from origin = 900 + 1350m = 2250m.
 
HENCE, THE TRAIN IS AT 225m and 2250m from its starting point, at half its maximum speed.
 
Thanks!