Let V be the speed of the bus running between towns A and B.
Given,
Speed of the cyclist = 20 km/hr
Relative speed of bus moving in the direction of the cyclist = V - v = (V-20) km/hr
Every 18 mins, the bus went past the cyclist. moving in the direction of the bus.
That is,
Distance covered by the bus =
One bus leaves every t minutes.
Therefore,
Distance travelled by the bus = V ... (ii)
Equations (i) and (ii) are equal.
Relative speed of the bus moving in the opposite direction of the cyclist = (V + 20) km/h
Time taken by the bus to go past the cyclist = 6 min = 6 / 60 h
Therefore,
Now, from equations (3) and (4), we get
In case of train A:
Initial velocity, u = 72 km/h = 20 m/s
Time, t = 50 s
Since the train is moving with uniform velocity,
Acceleration, aI = 0
Now, using second equation of motion,
Distance covered by train A is given by,
For train B,
Initial velocity, uB = 72 km/h = 20 m/s
Acceleration, a = 1 m/s2
Time, t = 50 s
Now, using second equation of motion,
Distance travelled by second train B is,
Length of both trains = 2
Therefore, the original distance between the driver of train A and the guard of train B = 2250 - 1000 - 800
= 450m.