$\underline{\text { Folklore }}$

A person descends a stationary escalator faster than they ascend it. Which is faster: to go down and then up a moving upwards escalator, or to go down and then up a moving downwards escalator?

(It is assumed that all speeds mentioned are constant, with the escalator moving at the same speed upwards and downwards, and the person's speed relative to the escalator is always greater than the speed of the escalator.)
Let the speed of the escalator be denoted by $v$, the speeds of the person ascending and descending the escalator by $v_{1}$ and $v_{2}$ respectively (then

$v<v_{1}<v_{2}$). Let the length of the escalator be denoted by $l$. The time taken to move on the ascending escalator is

$t_{1}=\frac{1}{v_{1}+v}-\frac{1}{v_{2}-v}$, and on the descending escalator $-t_{2}=\frac{1}{v_{2}+v}-\frac{1}{v_{1}-v}$.

We compute the difference $t_{1}-t_{2}=2 l v\left(\frac{1}{v_{2}^{2}-v^{2}}-\frac{1}{v_{1}^{2}-v^{2}}\right)$. According to the condition, the denominators of both fractions are positive, and the first is greater than the second, so the entire expression is negative. Conclusion: $t_{1}<t_{2}$.

## Answer

On the ascending escalator.