Q. 93.7( 113 Votes )

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 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 :

Given,


Speed of the cyclist, v = 20 km/h = 5.55 m/s


Time period for bus moving in the same direction= 18 min = 1080 s


Time period for bus moving in opposite direction=6 min = 360 s


Let, The velocity if the buses be V.


Thus,


Relative velocity of bus moving in the direction of cyclist = (V-5.55) m/s


Relative velocity of bus moving in opposite direction to the cyclist,


= (V+5.55) m/s


Distance covered by same direction bus = (V-5.55)×1080 m…….(1)


Distance covered by opposite direction bus = (V+5.55)×360 m…..(2)


Since both buses cover same distance (VT), equations (1) and (2) are equal.


(V-5.55)×1080 m = (V+5.55)×360 m


V = 11.11 m/s


Thus,


Equating, equation (1) = VT


(11.11-5.55)×1080 = 11.11×T


We get, T = 540 s

The buses move with a speed of 11.11 m/s or 40kmph and the ply after every 540 s or 9 mins

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