Question

Two trains a and b are connected by a regular bus service with a bus leaving in either direction every t min. a man cycling with a speed of 20 km/h in the direction a to b notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. what is the period t of the bus service and with what speed (assumed constant) do the buses ply on the road.

Answer 1

The Distance between two buses just when one takes off is V.T
where V is the velocity of the bus and T is frequency of the bus.
U is the speed of the cyclist=20km/hr


Note that when they move in the same direction, Relative velocity=V-U
The time to cover the lag of VT distance (that separates the two buses) is t=18 mins= 18/60 hrs.
Distance= t.(V-U)
This will be equal to V.T above

Hence, t(V-U) = VT  =>  VT=3/10.(V-U)

Now both

Also, Note that when they move in the opposite direction, Relative velocity=V+U

The time to cover the lag of VT distance (that separates the two buses) is t=6 mins= 6/60 hrs.

again you can equate them to get t(V+U) = VT 
=> VT=1/10(V+U)


equating VT from the above 2, 
3/10.(V-U) = 1/10(V+U)

3V-3U=V+U
2V=4U
V=2U

V=40Km/hr

Substituting, 
40.T=1/10(40+20)

T=3/20 hrs, T=9 minutes

Answer 2

Please refer to the above picture