Escalator Problem?

Question

You walk upwards on an escalator, with a speed of 1 step per second. After 50 steps you are at the end. You turn around and run downwards with a speed of 5 steps per second. After 125 steps you are back at the beginning of the escalator. How many steps do you need if the escalator stands still?

Answer 1

Let S be the escalator's speed in steps/second. Going up the escalator takes 50 seconds and going down takes 25 seconds (125 steps / 5 steps/second). Thus, 50*(1+S)=25*(5-S). It follows then that S=1. Therefore, the total number of steps traversed is 50*(1+1) or 100 steps if the escalator stands still.

Answer 2

Let's consider V the speed of the escalator. When going up, V is added to your speed (because it helps you get up faster) and when you go down it's subtracted (because it's trying to take you up, while you're trying to get down). Speed is distance over time. Even if you go up or down, the distance is the same because the escalator has the same length. So:

V = D / t => D = V * t

(V + 1 step/sec) * 50 = (5 steps/sec - V) * 625 = D (distance)

When you get down, your speed it's higher than the escalator's because you are eventually reaching the bottom floor. If V would have been higher, you would still be going up at the speed of 5 steps/sec. That's why you subtract V from your speed and not your speed from V instead.

When going up, it takes you 50 seconds. When going down, it takes you 25 seconds. (125 / 5 = 25)

(V + 1 step/sec) * 50 = (5 steps/sec - V) * 25 <=> 50*V + 50 steps/sec = 125 steps/sec - 25*V <=> 75*V - 75 steps/sec = 0 <=> V = 1 step/sec

D = (V + 1 step/sec) * 50 <=> D = 2 steps/sec * 50 sec = 100 steps if the elevator is still.

Answer 3

100 steps
Let the speed of the escalator =s steps per second
50/(s+1) = 150/(s+5)
3s +3 =s+5
2s =2 or s=1
Speed while going up = 2 steps per second and moved 50 steps, but actually he has moved 100 steps if the escalator is stand still

Answer 4

The escalator is moving with a velocity of v steps/ sec. The distance up the escalator is 50 steps + vt . Since your speed is 1 step/sec and you go 50 steps, going up t=50sec. Going down the distance you go is 125steps- vt, where t is the time it takes to run down. You are running 5steps/sec going 125 steps, so t=25sec. Since the distance up and down is the same, 50+v(50sec)=125-v(25sec). Group the like terms to get 125-50=v(50+25 sec) or 75steps=v(75sec) . Solve for v to get v= 1step/sec. So how far is is up the escalator? d=50steps +1step/sec(50sec) =50steps+50steps=100steps. So if the escalator were stopped, you would take 200steps, 100 up and 100 down. Notice that you can check your work by seeing that going down the distance is 125steps-v(25secs)=125-25=100steps, which is the same distance you went going up.