## Mathematics Puzzle: Round-trip On A Treadmill

Alice and Basil are bored while waiting in an airport. They decide to play a game with one of the nearby “moving sidewalks” (treadmills).

They decide that Alice will walk on the treadmill at constant speed v (with respect to the treadmill) for a distance y, and then turn around and walk back to the beginning at the same constant speed (with respect to the treadmill). Meanwhile, Basil will walk on the floor next to the treadmill at a constant speed v (with respect to the floor) for a distance y, and then turn around and come back to the starting point at the same constant speed (with respect to the floor).

The treadmill is much longer than the distance y. And the treadmill moves at a constant speed x.

Who wins this race?

(My apologies for not giving the source of the problem yet; I shall do so once I post some solutions in a day or two.)

I have taught mathematics and physics since the mid 1980s. I have also been a textbook writer/editor since then. Currently I am working independently on a number of writing and education projects while teaching physics at my local university. I love math and physics, and love teaching and writing about them. My blog also discusses education, science, environment, etc. https://qedinsight.wordpress.com Further resources, and online tutoring, can be found at my other site http://www.qedinfinity.com
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### 9 Responses to Mathematics Puzzle: Round-trip On A Treadmill

1. Well, I’m going to make a few assumptions here. First, I’ll assume that the treadmill is moving at some constant velocity, say a and we have 0<a<v. Now by observation, it really should matter what the direction the treadmill is going so let's just assume that when Alice walks her first y distance, the treadmill is moving in her direction. Based on my calculations, I got that Alice will beat Basil's time by $v^{2}y - 2ay +a^{2}$ .

• Or rather it should be $v^{2}y-2av+a^{2}$.

• Hi Stochastic Seeker,

Thanks for mentioning the assumption that the treadmill goes at a constant speed … I’ll update the text of the problem.

All the best,
Santo

2. Okay, it seems that I did make an arithmetic error in my calculations, and redoing them from scratch I got… Alice wins by a time of (forgive my latex errors, if there are any): $\frac{2ya^{2}}{v(v+a)(v-a)}$ seconds. I think the dimensions should be correct and when I do try out a=0, then the time difference does indeed go to zero. If it is correct, I’ll post my answer up upon request.

• Correction here. Basil should be the one who wins. As for why he intuitively wins, I will have to think more on that one.

On another note, this question seems to be very similar to a question that I did with projectile motion modeled with air resistance where the force of the air resistance acts in the opposite direction of the driving force and is proportional to the speed of the projectile.

3. So here’s my solution to the problem:

By observation, we can see that Basil will take a total of time $\frac{2y}{v}$ seconds. For Alice, we can see that she will take $\frac{y}{v+x}+\frac{y}{v-x}=\frac{vy-xy+vy+xy}{(v+x)(v-x)}=\frac{2v^{2}y}{v(v+x)(v-x)}$ seconds. We can rewrite Basil’s time as follows: $\frac{2y}{v}=\frac{2y(v+x)(v-x)}{v(v+x)(v-x)}=\frac{2vy^{2}-2vx^{2}}{v(v+x)(v-x)}=\frac{2vy^{2}}{v(v+x)(v-x)}-\frac{2vx^{2}}{v(v+x)(v-x)}$ . Thus, Basil will will by a time of $\frac{2vx^{2}}{v(v+x)(v-x)}$ seconds.

• Nick says:

We can write:
Basil’s time = 2yv / (v^2) seconds, and
Alice’s time = y/(v + x) + y/(v – x) = 2yv / (v^2 – x^2) seconds.

Looking at the two expressions for time, Alice’s time is greater because, while the numerators are the same, Alice’s denominator is smaller. (We assume v > x, of course!)

That Basil will win is implied by the fact that the graph of the function y = 1/x (where x and y have their usual meaning in this context; i.e., not as above) is concave for x > 0.

• Thanks to both Stochastic Seeker and Nick for their insights, which I will include in the body of the post once I update it with a solution.

Nick, the concave-up insight is welcome, and it will be nice to supplement it with a physical interpretation.

• Nick says:

An intuitive way to approach the problem is to note that Alice travels faster than Basil on the outbound leg by the same speed differential as she travels slower than him on the return leg. However, since Alice takes *longer* to complete the return leg than she does the outbound leg, her negative speed differential acts for longer than her positive speed differential, so, overall, she takes longer.

I’ve seen this same puzzle in the guise of two planes making a return trip from A to B. Plane 1 makes the return trip in still air. When plane 2 makes the trip, there is a constant wind blowing from A to B, so that for the outbound trip there is a tailwind and for the return trip there is a headwind. Which plane takes longest?