At a metro station, a girl walks up a stationary escalator in time $t_{1}$ . If she remains stationary on the escalator, then the escalator take her up in time $t_{2}$ . The time taken by her to walk up on the moving escalator will be

\frac{(t_{1} + t_{2})}{2}

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\frac{t_{1} t_{2}}{(t_{2} - t_{1})}

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\frac{t_{1} t_{2}}{(t_{2} + t_{1})}

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t_{1} - t_{2}

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Solution

The correct option is D $\frac{t_{1} t_{2}}{(t_{2} - t_{1})}$
Let speed of girl on stationary escalator $= V_{g} = (V_{1})$
speed of only escalator $= V_{e} = (V_{2})$
slant distance $= d$
Now, time to go up when escalator is stop, $t_{g} = \frac{d}{V_{g}} = ({= t}_{1})$
$\Rightarrow V_{g} = \frac{d}{t_{g}}$
time req. to go-up when girl stops & escalator moves, $t_{e} = \frac{d}{V_{e}} ({= t}_{2})$
$\Rightarrow V_{e} = \frac{d}{t_{2}}$
Now, when both girl & escalator moves up,velocity of girl with respect ot ground $= V_{g} + V_{e}$
Time taken, $t = \frac{d}{V_{g} + V_{e}}$

$= \frac{d}{\frac{d}{t_{1}} + \frac{d}{t_{2}}}$

$t = \frac{t_{1} t_{2}}{t_{2} + t_{1}}$

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