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?

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Vishal Baghel | Contributor-Level 10

5 months ago

3.9 Cycling speed $V_{c}$ = 20 kmph

Let as assume speed of the bus is $V_{b}$

The relative velocity of the buses in the direction of motion of the cyclist = $V_{b}$ - $V_{c}$

Buses go past him every 18 mins = 18/60h = 0.3h

Distance covered = ( $V_{b} - V_{c}$ ) * 0.3

Since the buses leave every T minutes, ( $V_{b} - V_{c}$ ) * 0.3 = $V_{b}$ * (T/60) ……(1)

The relative velocity of the buses in the opposite direction of the cyclist = $V_{b}$ + $V_{c}$

Buses go past him in the opposite direction = every 6 m = 6/60 h

Distance covered = ( $V_{b}$ + $V_{c}$ ) * 1/10

Since the buses leave every T minute, we get ( $V_{b}$ + $V_{c}$ ) * 1/10

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3.9 Cycling speed $V_{c}$ = 20 kmph

Let as assume speed of the bus is $V_{b}$

The relative velocity of the buses in the direction of motion of the cyclist = $V_{b}$ - $V_{c}$

Buses go past him every 18 mins = 18/60h = 0.3h

Distance covered = ( $V_{b} - V_{c}$ ) * 0.3

Since the buses leave every T minutes, ( $V_{b} - V_{c}$ ) * 0.3 = $V_{b}$ * (T/60) ……(1)

The relative velocity of the buses in the opposite direction of the cyclist = $V_{b}$ + $V_{c}$

Buses go past him in the opposite direction = every 6 m = 6/60 h

Distance covered = ( $V_{b}$ + $V_{c}$ ) * 1/10

Since the buses leave every T minute, we get ( $V_{b}$ + $V_{c}$ ) * 1/10 = $V_{b}$ * T/60 …….(2)

From equations (1) & (2), we get

( $V_{b}$ + $V_{c}$ ) / 10 = ( $V_{b} - V_{c}$ ) * 0.3

$V_{b}$ + $V_{c}$ = 3( $V_{b}$ - 3 $V_{c}$ )

2 $V_{b}$ = 4 $V_{c}$ , $V_{b}$ = 2 $V_{c}$ So $V_{b}$ = 2 * 20 = 40 kmph

From the equation (1), we get

(40-20) * 0.3 = 40 * T/60, T = 9 min

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