A boy reaches the airport and finds that the escalator is not working. He walks up the stationary escalator in time $t_{1}$ . If he remains stationary on a moving escalator, then the escalator takes him up in time $t_{2}$ . The time taken by him to walk up on the moving escalator will be:

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

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

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

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t_{2} - t_{1}

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Solution

The correct option is C $\frac{t_{1} t_{2}}{t_{2} + t_{1}}$
Let $L$ be the length of the escalator.

When only boy is walking,

$v_{b} = \frac{L}{t_{1}}$

And when only escalator is moving,

$v_{e s c} = \frac{L}{t_{2}}$

Now let $t$ be the time taken by boy to walk up on the moving escalator.

$∴ t = \frac{L}{v_{b} + v_{e s c}} = \frac{L}{(L / t_{1}) + (L / t_{e s c})}$

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

Hence, option $(C)$ is correct.

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A boy reaches the airport and finds that the escalator is not working. He walks up the stationary escalator in time t1. If he remains stationary on a moving escalator, then the escalator takes him up in time t2. The time taken by him to walk up on the moving escalator will be:

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