A passenger reaches the platform and finds that the second last boggy of the train is passing him. The second last boggy takes 3 sec to pass the passenger and the last boggy takes 2 sec to pass him. Find the time by which the passenger late for the departure of the train? Assume that the train accelerates at constant rate and all the boogies are of equal length.

Let say Velocity of Train = V  at the time of 2nd last boggie start crossing Man
Distance Covered in 3 sec
S = V3  + (1/2)a3²
S = 3V + 4.5a
Distance Covered in 2 secs
Distance Covered in 5 sec - Distance Covered in 3 sec
S = 5V + 12.5a - 3V - 4.5a  = 2V +  8a
as length of boggies are equal
3V + 4.5a = 2V +  8a
=> V = 3.5a
V = 0 + at
=> 3.5a = 0 + at
=> t = 3.5
passenger is 3.5 sec late for the departure of the train​