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

Let the length of escalator be L.

Also, suppose that the velocity of boy with respect to escalator be v₁ and the velocity of escalator be v₂.

$\therefore$ Velocity of boy with respect to stationary escalator can be given as

${\mathrm{v}}_1=\frac{\mathrm{L}}{{\mathrm{t}}_1}$                           ...(i)

When escalator is moving and boy is at rest velocity of escalator can be given as

${\mathrm{v}}_2=\frac{\mathrm{L}}{{\mathrm{t}}_2}$                        ...(ii)

So, the resultant velocity can be given as when both escalator and boy are moving,

$\mathrm{v}={\mathrm{v}}_1+{\mathrm{v}}_2$                 ...(iii)

From Eqs. (i), (ii) and (iii), we get

$\begin{array}{l}\mathrm{v}=\frac{\mathrm{L}}{{\mathrm{t}}_1}+\frac{\mathrm{L}}{{\mathrm{t}}_2}\Rightarrow \frac{\mathrm{v}}{\mathrm{L}}=\frac{1}{{\mathrm{t}}_1}+\frac{1}{{\mathrm{t}}_2}\\ \Rightarrow \frac{\mathrm{v}}{\mathrm{L}}=\frac{{\mathrm{t}}_1+{\mathrm{t}}_2}{{\mathrm{t}}_1{\mathrm{t}}_2}\Rightarrow \frac{\mathrm{L}}{\mathrm{v}}=\frac{{\mathrm{t}}_1{\mathrm{t}}_2}{{\mathrm{t}}_1+{\mathrm{t}}_2}\\ \Rightarrow \mathrm{t}=\frac{{\mathrm{t}}_1{\mathrm{t}}_2}{{\mathrm{t}}_1+{\mathrm{t}}_2} [\because \mathrm{t}=\mathrm{L}/\mathrm{v}]\end{array}$