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?
Let V be the speed of the bus running between towns A and B.
Speed of the cyclist, v = 20 km/h
Relative speed of the bus moving in the direction of the cyclist
= V – v = (V – 20) km/h
The bus went past the cyclist every 18 min i.e. 18/60 h, (when he moves in the direction of the bus).
Distance covered by the bus = (V - 20) 18/60………….. (i)
Since one bus leaves after every T minutes, the distance travelled by the bus will be equal to:
V x T/60 .............(ii)
Both equations (i) and (ii) are equal.
(V - 20) x 18/60 = VT / 60 .............. (iii)
Relative speed of the bus moving in the opposite direction of the cyclist = (V + 20) km/h
Time taken by the bus to go past the cyclist = 6 min = 6/60 h
∴(V - 20) x 6/60 = VT / 60 ................ (iv)
From equations (iii) and (iv), we get
(V - 20) x 6/60 = (V - 20) x 18/60
V - 20 = 3V - 60
2V = 80
V = 40 km/h
Substituting the value of V in equation (iv), we get
(40+20) x 6/60 = 40T/60
60 x 6/60 = 40T/60
6 = 40T/60
⇒ 40T = 6x60
⇒ 40T = 360
⇒ T = 360/40
⇒ T = 9 min
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