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?
I'm thinking, if the first person was traveling three steps per stride (say) then both people could be traveling down an up escalator. Again using Tristans starting point...
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
Now the time function has changed for the long legged man
the times are 150/m and 75/(3m) respectively
150 - v(150/m) = s
75 - v(75/3m) = s
v/m = 1-s/150
75 - 25(1-s/150) = s
s = 60 steps
I think there are other solutions for the various combinations of stride length and escalator direction.
Posted by Percy
on 2006-04-01 20:33:17