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Relative Motion in 1D

On a linear e...

Question

On a linear escalator running between two points A and B, a boy takes time $t_{1}$ to move from A to B, if the boy runs with a constant speed on the escalator. If the boy runs from B to A, he takes time $t_{2}$ to reach B from A. What is the time taken by the boy to move from A to B if he stands still on the escalator?
[The escalator moves from A to B]

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Solution

Let the speed of escalator be $V_{e}$
Distance traveled =S(from A to B)
$t_{1} = \frac{S}{V_{b} + V_{e}}$
$t_{2} = \frac{S}{V_{b} - V_{e}}$
If he stands on escalator, $t = t i m e t a k e n = \frac{S}{V_{e}}$
$\frac{t_{1}}{t_{2}} = \frac{V_{b} - V_{e}}{V_{b} + V_{e}} t_{1} V_{b} + t_{1} V_{e} = t_{2} V_{b} - t_{2} V_{e} (t_{1} + t_{2}) V_{e} = (t_{2} - t_{1}) V_{b}$
$V_{b} = (\frac{t_{1} + t_{2}}{t_{2} - t_{1}}) V_{e} t_{1} = \frac{S}{V_{e} + (\frac{t_{1} + t_{2}}{t_{2} - t_{1}}) V_{e}} = \frac{S}{V_{e} [\frac{t_{2} - t_{1} + t_{2} + t_{1}}{t_{2} - t_{1}}]} = \frac{S}{V_{e} (\frac{2 t_{2}}{t_{2} - t_{1}})} t = \frac{2 t_{1} t_{2}}{t_{2} - t_{1}}$

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