Ncert Exercises & Solutions

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?

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 (18/60h)

Distance covered by the bus = $(V - 20) \frac{18}{60} k m . . . . (1)$ $(V - 20) \frac{18}{60} k m . . . . (1)$

Since one bus leaves after every T minutes, the distance traveled by bus will be equal to:

$V \times \frac{T}{60} . . . . . (2)$ $V \times \frac{T}{60} . . . . . (2)$

Both equations (1) and (2) are equal.

(V - 20) \times \frac{18}{60} = \frac{V T}{60} . . . . . (3)

$(V - 20) \times \frac{18}{60} = \frac{V T}{60} . . . . . (3)$

The relative speed of the bus moving in the opposite direction of the cyclist = (V + 20) km/h

Time is taken by the bus to go past the cyclist= $6 m i n = \frac{6}{60} h$ $6 m i n = \frac{6}{60} h$

$\Rightarrow (V + 20) \frac{6}{60} = \frac{V T}{60} . . . . . (4)$ $\Rightarrow (V + 20) \frac{6}{60} = \frac{V T}{60} . . . . . (4)$

From equations (3) and (4), we get

$(V + 20) \times \frac{6}{60} = (V - 20) \times \frac{18}{60}$ $(V + 20) \times \frac{6}{60} = (V - 20) \times \frac{18}{60}$
$V + 20 = 3 V - 60$ $V + 20 = 3 V - 60$
$2 V = 80$ $2 V = 80$
$V = 40 k m / h$ $V = 40 k m / h$

Substituting the value of V in equation (4), we get

$(40 + 20) \times \frac{6}{60} = \frac{40 T}{60}$ $(40 + 20) \times \frac{6}{60} = \frac{40 T}{60}$
$T = \frac{360}{40} = 9 m i n$ $T = \frac{360}{40} = 9 m i n$

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