When a person walks down the escalator, which goes down, he spends 1 minute on the descent.

When a person walks down the escalator, which goes down, he spends 1 minute on the descent. If he walked 2 times faster, he would descend in 45 seconds, indicate how long (in minutes) a person will descend while standing, on a movable escalator.

Given:

t1 = 1 minute = 60 seconds – the time during which a person goes down the escalator going down;

t2 = 45 seconds – the time during which a person would descend if he moved 2 times faster.

It is required to determine t (seconds) – the time of descent of a motionless person along the escalator.

Let the length of the escalator be L, the speed of the person v, and the speed of the escalator v1. Then we have 2 equations:

L / (v + v1) = t1 (1);

L / (2 * v + v1) = t2 (2);

From the first equation we find the speed of a person v:

L = t1 * (v + v1);

L = t1 * v + t1 * v1;

t1 * v = L – t1 * v1;

v = (L – t1 * v1) / t1.

We substitute the found speed value into the second equation:

L / (2 * v + v1) = t2;

L = 2 * v * t2 + v1 * t2;

L = 2 * t2 * (L – t1 * v1) / t1 + v1 * t2;

L * t1 = 2 * t2 * (L – t1 * v1) + v1 * t2 * t1;

L * t1 = 2 * L * t2 – 2 * v1 * t2 * t1;

L * t1 = 2 * L * t2 – v1 * t1 * t2;

v1 * t1 * t2 = 2 * L * t2 – L * t1;

v1 * t1 * t2 = L * (2 * t2 – t1);

v1 = L * (2 * t2 – t1) / (t1 * t2).

Knowing the speed of the escalator, we find the descent time of a motionless person:

t = L / v1 = L / (L * (2 * t2 – t1) / (t1 * t2)) = t1 * t2 / (2 * t2 – t1) =

= 60 * 45 / (2 * 45 – 60) = 2700 / (90 – 60) = 2700/30 = 90 seconds.

Answer: a person will descend in 90 seconds.