NCERT solution Class 11 motion in a straight line

NCERT Solutions

Class 11 Physics

Motion In A Straight Line

NCERT Questions

Q. 2.1
Q. 2.2
Q. 2.3
Q. 2.4
Q. 2.5
Q. 2.9
Q. 2.6
Q. 2.7
Q. 2.8
Q. 2.10
Q. 2.11
Q. 2.12
Q. 2.13
Q. 2.14
Q. 2.15
Q. 2.16
Q. 2.17
Q. 2.18

Q. 2.9

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?

Let V be the speed of the bus running between towns A and B.

Speed of the cyclist, v_c = 20 km/h

Case 1:- Relative speed of the bus moving in the direction of the cyclist

= (V – v_c) = (V – 20) km/h

Since the bus goes past the cyclist every 18 min i.e., (18/60),

Therefore, the distance covered by the bus w.r.t the cyclist

= (V – 20) x (18/60) km (i)

As the bus leaves after every T minutes, the distance travelled by the bus will be

= (V x (T/60)) (ii)

As both equations (i) and (ii) are equal.

= (V – 20) x (18/60) = (VT)/ (60) (iii)

Case 2:- Relative speed of the bus coming from town q to p w.r.t cyclist

= (V+20) kmh^-1

Time taken by the bus to go past cyclist = 6 min = (6/60) h

Therefore (V +20) (6/60) = (VT)/ (60) (iv)

From equations (iii) and (IV)

(V+20) (6/60) = (V – 20) x (18/60)

(V + 20) = 3V – 60

2V = 80

V = 40km/h

Substituting the value of V in equation (iv), we get

(40 + 20) x (6/60) = (40T)/ (60)

T = (360/40) = 9min

Video solution:

Motion In A Straight Line Class 11 Physics NCERT Chapter 3 NCERT Solutions

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NCERT solutions of Motion In A Straight Line Class 11 Physics - all questions of all exercises are solved. Q. No. 1, 2, 3, and so on. Complete explanation and detailed solution makes them the best videos for Class 11 Physics Motion In A Straight Line NCERT solutions.