__Question 3.9__

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?

__Solution:__

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) X 18 / 60 km .... (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

T = 360 / 40 = 9 min

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