Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km h�1 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?

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