Two cities A and B are connected by a regular bus services with buses plying in either direction every T seconds. The speed of each bus is uniform and equal to $V_{b}$ . A cyclist cycles from A to B with a uniform speed of $V_{c}$ . A bus goes past the cyclist in every $T_{1}$ second in the direction A to B and every $T_{2}$ second in direction B to A. Then

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$T_{2} = \frac{V_{b} T}{V_{b} + V_{c}}$

$T_{1} = \frac{V_{b} T}{V_{b} + V_{c}}$

$T_{1} = \frac{V_{b} T}{V_{b} - V_{c}}$

$T_{2} = \frac{V_{b} T}{V_{b} - V_{c}}$

answer is C, D.

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Detailed Solution

Distance between two buses on road $= V_{b} T$
For A to B direction: distance = relative velocity x time
$V_{b} T = (V_{b} - V_{c}) T_{1} \Rightarrow T_{1} = \frac{V_{b} T}{V_{b} - V_{c}}$
For B to A direction: $V_{b} T = (V_{b} + V_{c}) T_{2} \Rightarrow T_{2} = \frac{V_{b} T}{V_{b} + V_{c}}$

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