Two people running up and down escalators

Question

Couldn't really figure out a good title for this post, I am sorry.

But here's the problem:

PREMISE: Two people, A and B, are running, with speed u, up and down separate escalators with length L.

B's escalator is running with speed v, but A's escalator does not work.

QUESTION: Would they arrive at the same time? What would happen?

My attempt:

$$T_A = \frac{2L}{u} $$ $$T_B = T_B1 + T_B2 = \frac{L}{u+v} + \frac{L}{u-v} = \frac{2uL}{u^2 - v^2}$$ Here is where I am a bit uncertain if I can do this, I assign a factor F to find a relation between the two expressions: $$T_A * F = T_B$$ $$\frac{2L}{u} * F = \frac{2uL}{u^2 - v^2}$$ $$F = \frac{u^2}{u^2 - v^2}$$ If F > 1, that would mean that T_B is greater, and if F < 1, that would mean that T_A is greater.

$$u^2 - v^2 > u^2 $$ .. doesn't hold since that would mean that $v^2$ must be negative therefore $$u^2 - v^2 <= u^2 $$ holds which would give $F >= 1$ which implies T_B is greater and therefore T_A wins everytime.

Does my math check out, can I make the assumptions that I make? And also, if I am right, is there an easy way to see why this is? It seems really counter-intuitive to me.

Answer 1

Your math looks correct to me.

You can define $F$ as you have, the reason being that $T_A \cdot F=T_B$ if and only if $F=\frac{T_B}{T_A}$ (provided $T_A \neq 0$). So you can simply define $F$ to be $\frac{T_B}{T_A}$. As you've shown, this quantity must always be $\geq 1$, so $A$ always wins.

To see intuitively why $A$ should win, it might help to think about the case when $v$ is really close to $u$. Suppose $u=2$, $v=1.9$, and $L=10$. Then when $B$ is moving against the escalator, $B$ has a net speed of $2-1.9=0.1$. It will take $\frac{L}{v-u}=\frac{10}{0.1}=100$ units of time to go against the escalator. Meanwhile it takes $A$ only $\frac{L}{u}=\frac{10}{2}=5$ units of time both ways. It doesn't matter how much time $B$ saved moving with the escalator, because it will take much longer to move against the escalator than it will take $A$ to go up and down. Of course this isn't necessarily the case when $u$ and $v$ are not so close, but hopefully the example illustrates why $B$ should not expect to have the advantage.