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 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
# Answer-
T = 9 min
V = 40 km/h
# Explaination-
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 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
Thus, buses ply on the road each 9 minutes and speed of the bus is 40 kmph.
Hope that is helpful...
Given:
Let the speed of the bus be V that is running between town A and B.
Speed of the cyclist that is given = 20 km/hr
Relative speed of the bus that is moving in the direction of the cyclist
= (V - v) = (V - 20) km/hr
The bus went past the cyclist every 18 mins this means 18/60 hrs
(This happens when he moves in the direction of the bus)
Solving:
The total distance that is covered by the bus = (V - 20) 18 / 60 km ----Eq(i)
Since one bus leaves after every T mins, the distance which is traveled by the bus is equal to:
V x T / 60 -------Eq(ii)
Both the equations (i) and (ii) are equal:
(V - 20) x 18/60 = VT / 60 -------Eq(iii)
The relative speed of the bus that is moving in the opposite direction of the cyclist = (V + 20) km/hr
The Time taken by the bus to go past the cyclist
= 6 mins
Converting into hours we get = 6 / 60 hrs
Therefore, (V + 20) 6 / 60 = VT / 60-------Eq(iv)
From equation (iii) and equation (iv) we get:
(V + 20) x 6 / 60 = (V - 20) x 18 / 60
V + 20 = 3V - 60
2V = 80
V = 80 / 2
V = 40 km/hr
Substituting the value of V in equation (iv), we get:
(40 + 20) x 6 / 60 = 40 T / 60
(60) x 6/60 = 40 T / 60
T = 360 / 40
T = 9 mins
Therefore, buses ply on the road for 9 minutes and speed of the buses is 40 kilometre per hour.