# A person walking down the descending escalator spends 1 minute on the descent.

A person walking down the descending escalator spends 1 minute on the descent. If a person walks twice as fast, he will spend 15 seconds. less. How long will he descend while standing on the escalator?

Given:

t1 = 1 minute = 60 seconds – the time spent by a person walking down the escalator;

t2 = t1 – 15 = 45 seconds – if a person’s speed doubles, he will descend 15 seconds faster.

It is required to determine t (seconds) – how much time a person will spend if he stands motionless on the escalator.

Let the speed of a person be v1, the speed of the escalator is v2, and the length of the escalator is l. Then we have two equations:

l = (v1 + v2) * t1 (1);

l = (2 * v1 + v2) * t2 (2).

From equation (1) we find that:

v1 = l / t1 – v2.

Substituting the found value v1 into equation (2), we obtain:

l = (2 * l / t1 – 2 * v2 + v2) * t2;

l = (2 * l / t1 – v2) * t2;

l / t2 = 2 * l / t1 – v2;

v2 = 2 * l / t1 – l / t2 = l * (2 * t2 – t1) / (t1 * t2).

Then the time will be equal to:

t = l / v2 = l / (l * (2 * t2 – t1) / (t1 * t2)) = t1 * t2 / (2 * t2 – t1);

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

Answer: the time of descent of a standing person along the escalator will be equal to 90 seconds.

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