Two trains are running in the same directions.Train A is ahead of train B by 1 km at t=0. The velocity equation of train A is given by V = 10t . The acceleration equation for train B is a = 45 . After some time they meet at a station and a person gets down from B to meet a friend in A.The station is 61000m from starting point of train B.The person had to wait for some time,so that the train carrying his friend arrives at the station, where it is instantaneously brought to stop.Calculate the time he had to wait.

1) 52.068 s.
2) 12.234 s.
3) 25.489 s.
4) 67.328 s.

Train A-
The distance travelled by train A is sa = 60000 m.
The velocity equation is given by v=10t.



Train B-
The distance travelled by train B is sb = 61000 m.
The acceleration is given as 45 m/s2.
The velocity is


The time the person had to wait is
109.544 - 52.068 = 57.476 s.