Consider two cities P and Q between which consisted bus service is available in both the directions every 'x' minutes. A morning jogger is jogging towards Q from P with a speed of 10 km/hr. Every 18 minutes a bus crosses this jogger in its own direction of motion and every 6 minutes another bus crosses in opposite direction. What is the time between two consecutive buses and also find the speed of the buses?

6 min, 15 km/hr

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9 min, 15 km/hr

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6 min, 20 km/hr

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9 min, 20 km/hr

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Solution

The correct option is D
9 min, 20 km/hr

Analysis of Problem:
The equations of motion are applicable in any frame of reference. Attaching the frame of reference at jogger. The velocity of bus is now required w.r.t. jogger. Suppose speed of buses = v km/h
The relative velocity of bus w.r.t. the jogger,
While moving from P to Q = (v - 10) km/h
While moving from Q to P = (v + 10) km/h
So the distance travelled by bus in x minutes = $\frac{D i s t a n c e}{S p e e d} = T i m e \Rightarrow x . v$
In case of motion from P to Q = $\frac{v . x}{v - 10} = 18 \Rightarrow v . x = 18 v - 180 . . . . . (i)$

In case of motion from Q to P = $\frac{v . x}{v + 10} = 6 \Rightarrow v . x = 6 v + 60 . . . . . (i i)$

From comparing equations (i) and (ii), we find 18v-180=6v+60 $\Rightarrow$ 12v=240 $\Rightarrow$ v=20 km/hr
Using this value of v in (i), x =9 min

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