On a working escalator, one man walking down it took 50 steps. Another man, walking three times faster, took 75 steps.
How many steps would each man take going down the same escalator during a power outage?
Let v = the velocity of the escalator (down)
m = the velocity of the first man (down)
s = the number of steps on the escalator
The velocity of the second man is 3m
The amount of time for each man to go down is 50/m and 75/(3m) respectively
50 + v(50/m) = s
75 + v(75/3m) = s
v/m = s/50 - 1
75 + 25(s/50 - 1) = s
50 = s/2
s = 100 steps
That is, unless someone falls because it's so dark.
Posted by Tristan
on 2006-03-29 20:37:17