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 $20 k m / h$ 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?

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Solution

Let the speed of each bus = v km/h
The distance between the nearest buses plying on either car = vT km...(i)
For buses going from town A to B:
Relative speed of bus in the direction of motion of man = (v - 20)
Buses plying in this direction go past the cyclist after every 18 min. Therefore, separation between the buses
= (v - 20) $\times$ (18/60)
From (i), (v - 20) $\times$ (18/60) = vT ...(ii)
For buses coming from B to A:
The relative velocity of bus with respect to man = (v + 20) Buses coming from town B past'the cyclist after every 6 min,
$∴$ (v + 20) $\times$ (6/60) = vT ...(iii)
Solving (ii) and (iii), we get\
v = 40 km/h and T = 2/30 h

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