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? | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-3 Motion In A Straight Line



Q.9:- 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?

 

Answer:-



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
= Vv = (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 km    …. (i)
Since one bus leaves after every T minutes, the distance travelled by the bus will be equal to
V × T / 60    ….(ii)
Both equations (i) and (ii) are equal.
(V – 20) × 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) × 6 / 60 = VT / 60    ….(iv)
From equations (iii) and (iv), we get
(V + 20) × 6 / 60 = (V – 20) × 18 / 60
V + 20 = 3V – 60
2V = 80
V = 40 km/h
Substituting the value of V in equation (iv), we get
(40 + 20) × 6 / 60 = 40T / 60
T = 360 / 40 = 9 min