Xavier and Yonette are waiting for their plane at an airport, when Xavier proposes a race:
"See that moving sidewalk? Why don't you run on it, and then when you reach the end turn back and run to where you started from? I meanwhile will run the same path, but right next to it rather than on it."
Yonette thought about it and said:
"But I don't see the point... We both run with the same speed, and it's the same distance. Sure, I will gain a bit on you while the sidewalk increases my speed, but then I will just lose the advantage while I'm running back and we will arrive at the same time."
Assuming Xavier and Yonette do run with the same speed, who will win the race?
The answer has already been provided, but here are some simple (and probably redundant) equations that provide additional proof. Let D be the distance traveled in one direction, let S be the speed of each runner before consideration of the moving sidewalk, and let M be the speed of the moving sidewalk.
Therefore, the time required by Yonette is given as
D/(S + M) + D/(S - M) and the time required by Xavier is 2D/S.
This simplifies to ((DS -DM)+ DS + DM)/(S^2 - M^2) or
2DS/(S^2 - M^2) for Yonette vs 2D/S for Xavier.
We can multiply both the numerator and denominator of Xavier's computation by S to simplify the comparison.
This gives us 2DS/(S^2 - M^2) vs 2DS/S^2
Since the numerators of both equations are identical and the denominator of Yonnettes calculation must be less than the denominator of Xaviers (since M must be a positive number), Xavier will always require less time than Yonette. Gordon S.
For those who like equations
