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?

Answers (1)

Let the velocity of the bus be  V km h-1

Let the period of the bus be T minutes.

The speed of the bus travelling in the same direction as the cyclist relative to the cyclist is = (V-20) km h-1

The speed of the bus travelling in the opposite direction as the cyclist relative to the cyclist is = (V+20) km h-1

The distance between two consecutive buses travelling in the same direction is s

s=fracVT60 km

This distance s is in turn equal to the distance between the cyclist and the next bus at an instant when one bus goes past him.

fracVT60=frac(V-20)	imes 1860          (i)

fracVT60=frac(V+20)	imes 660           (ii)

Solving the above equations (i) and (ii) we get V=40 km h-1 and T=9 min

The buses travel at a speed 40 km h-1 and the period of the bus service is 9 minutes.

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