Let the train accelerates upto a distance of 'x'm and then decelerate and stop after (1215-x)m.
So at a point x , the velocity (v) is given by v^2 = 0 + 2x .
This velocityis retarted with a deceleration of 3m/s^2.
So 2x = 2*3*(1215-x).
This gives x = 911m.
So v^2 = 2x 911
v = 43 m/s.
Since, v = u + at ,
we get, 43= 0 + 1*t
t = 43 s
Now for the decelerating journey ,
0 = 43- 3*t
t =14 s.
So the total journey time = 43 + 14 = 57 seconds.