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Class 11
Physics
Motion in a straight Line
two towns a and b are connected by a regular bus s...

Question:
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

= 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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