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?
Given,
Speed of the cyclist, v = 20 km/h = 5.55 m/s
Time period for same direction bus pass by cyclist= 18 min = 1080 s
Time period for opposite direction bus pass by cyclist=6 min = 360 s
Let, The velocity if the busses be V.
Thus,
Relative velocity of bus moving in the direction 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