Question
Sun May 27, 2012 By: Anuj Kumar

# 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 hÂ–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?

Expert Reply
Wed May 30, 2012
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

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