13.252. The duty maintenance worker descended on a downward-moving metro escalator. His entire journey from the upper platform to the lower one lasted $24 \mathrm{s}$. Then he climbed up and at the same pace descended again, but this time on a stationary escalator. It is known that the descent lasted 42 s. How many seconds would it take for a person to descend on a moving downward escalator while standing on a step?
## Solution.

Let $l$ be the path along the stationary escalator, $v_{\text {esc. }}$ be the speed of the escalator, and $v_{\text {tech. }}$ be the speed of the technician along the stationary escalator.

Then $\frac{l}{v_{\text {tech. }}}=42, \frac{l}{v_{\text {tech. }}+v_{\text {esc. }}}=24$. We need to find $\frac{l}{v_{\text {esc. }}}$. We get

$$
\frac{v_{\text {tech. }}}{l}=\frac{1}{42}, \frac{v_{\text {tech. }}}{l}+\frac{v_{\text {esc. }}}{l}=\frac{1}{24} ; \frac{v_{\text {esc. }}}{l}=\frac{1}{24}-\frac{1}{42}=\frac{1}{56} \Rightarrow \frac{l}{v_{\text {esc. }}}=56 \mathrm{~s} .
$$

Answer: in 56 s.