Class 11 Physics - Chapter Motion in a straight Line NCERT Solutions | Two towns A and B are connected by a reg

Welcome to the NCERT Solutions for Class 11th Physics - Chapter Motion in a straight Line. This page offers a step-by-step solution to the specific question from Exercise 1, Question 9: two towns a and b are connected by a regular bus s....
Question 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

= 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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