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Standard XII

Physics

Relative Motion in 1D

Two towns A a...

Question

Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes with same speed. A man cycling with speed of 20km/h in the direction A to B, notices that a bus goes past him every $t_{1}$ =18 minutes in the direction of motion, and every $t_{2}$ =6 minutes in the opposite direction. What is the period T of the bus service? Assume that velocity of cyclist is less than velocity of bus.

4.5 minutes

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24 minutes

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9 minutes

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12 minutes

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Solution

The correct option is C 9 minutes
Distance cover by bicycle in each of bus pass is equal and therefore
$20 \times T = (20 + x) 6 = (20 - x) 18$
where $x =$ bus speed
$x = 10 m / m i n$
then
$20 \times T = 30 \times 6 T = 9 m i n u t e s$

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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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The correct option is C 9 minutesDistance cover by bicycle in each of bus pass is equal and therefore20×T=(20+x)6=(20−x)18where x= bus speedx=10m/minthen20×T=30×6T=9minutes

The correct option is C 9 minutes
Distance cover by bicycle in each of bus pass is equal and therefore
$20 \times T = (20 + x) 6 = (20 - x) 18$
where $x =$ bus speed
$x = 10 m / m i n$
then
$20 \times T = 30 \times 6 T = 9 m i n u t e s$