A boy reaches the airport and finds that the escalator

Question:
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 :

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

Correct Option: , 3

Solution:

$\mathrm{L}=$ Length of escalator

$\mathrm{V}_{\mathrm{b} / \mathrm{exc}}=\frac{\mathrm{L}}{\mathrm{t}_{1}}$

When only escalator is moving.

$V_{\text {esc }}=\frac{L}{t_{2}}$

when both are moving

$\mathrm{V}_{\mathrm{b} / \mathrm{g}}=\mathrm{V}_{\mathrm{b} / \mathrm{esc}}+\mathrm{V}_{\mathrm{esc}}$

$V_{b / g}=\frac{L}{t_{1}}+\frac{L}{t_{2}} \Rightarrow\left[t=\frac{L}{V_{b / g}}=\frac{t_{1} t_{2}}{t_{1}+t_{2}}\right]$

$\mathrm{V}_{\mathrm{b} / \mathrm{g}}=\frac{\mathrm{L}}{\mathrm{t}_{1}}+\frac{\mathrm{L}}{\mathrm{t}_{2}} \Rightarrow\left[\mathrm{t}=\frac{\mathrm{L}}{\mathrm{V}_{\mathrm{b} / \mathrm{g}}}=\frac{\mathrm{t}_{1} \mathrm{t}_{2}}{\mathrm{t}_{1}+\mathrm{t}_{2}}\right]$

A boy reaches the airport and finds that the escalator

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