0
$\begingroup$

Alice and Bob both live in point A. Both work in point B.

Every morning, the time it takes to get from A to B, starts at value X, goes up (with rush hour) to Y and then back to X (for our discussion, let's not define how it gets from X to Y and back but it is a continuous function).

Alice and Bob want to see if they can leave at different times in the morning and arrive at the same time. Assume that both drive the highest speed they can give the traffic, and no more than the speed limit.

So think about it this way: There is a function f that maps the departure time d (in seconds since midnight) to how long the trip would be in seconds (call it l for length). Is it possible for:

$$d_1+l_1=d_2+l_2$$

Can you prove if they can, or cannot? If they can, when?

For example, if the function is structured so that each second later that you depart, the trip gets shorter by a second, it's possible (essentially the function is behaving like $-d+b$ in that specific section). But can a function like this exist in the situation of traffic patterns?

$\endgroup$
5
  • $\begingroup$ "starts at X, goes up to Y" That's beyond vague. Also, what efforts have you tried to put into solving the problem yourself? $\endgroup$ – Don Thousand Nov 7 '18 at 1:08
  • $\begingroup$ I attempted the logical way: Bob would need to catch up to Alice if he leaves after. Any progress he makes to catch up to Alice, Alice will progress more. It's the typical infi problems (any time you progress, there's still more distance to cover). However, in those types of examples, you eventually catch up. So this seems theoretically possible $\endgroup$ – Y L Nov 7 '18 at 1:11
  • $\begingroup$ It's possible thanks to the intermediate value theorem: however, that doesn't change the fact that the question is still pretty poorly worded and vague (when X goes to Y, does it go linearly? exponentially? logarithmically?) and you need to add your thoughts to the question itself. $\endgroup$ – Don Thousand Nov 7 '18 at 1:13
  • $\begingroup$ Rushabh, do you disagree with the answer below? You said it IS possible. $\endgroup$ – Y L Nov 7 '18 at 1:45
  • $\begingroup$ I do agree it is possible. I just think the question is quite poor. $\endgroup$ – Don Thousand Nov 7 '18 at 2:32
1
$\begingroup$

Some assumptions. First, let's assume that they're taking the same path from $A$ to $B$. Now, let $x^\prime(t,x)$ be the velocity at time $t$ and position $x$. Now let's assume that $v$ is continuously differentiable with respect to $t$ and $x$ (which isn't a ridiculous assumption). Now, suppose that their paths intersect at time $t_1$ and position $x_1$. Then there must be multiple solutions to the initial value problem where $x^\prime(t)=v(t,x(t))$ and $x(t_1)=x_1$. However, that is impossible by the Picard–Lindelöf theorem, so the paths must there must never be any intersections between their paths.


With the more formal argument done, let's think about why this makes sense. Suppose that Alice and Bob are really close to each other. Then their velocities must be about the same as the derivative is continuous. In fact, the closer they get, the smaller the difference in velocity becomes. It so happens that the difference in velocity gets smaller fast enough that it would take an infinite amount of time for Bob to catch up. However, proving this is actually much more complicated.

$\endgroup$
2
  • $\begingroup$ So to summarize, you're saying that no matter how fast Bob is going in order to catch up with Alice, at some point he's right behind her and traveling at her speed and must arrive after her. Right? $\endgroup$ – Y L Nov 7 '18 at 1:43
  • $\begingroup$ @YoniL, basically yes. This wouldn't necessarily be true if the derivative weren't continuous (if for example, the speed limit changed and there were no traffic and so in our model, they instantly accelerated from one velocity to another. If there are speed limit changes, we have to assume there is no instant change in velocity (which is accurate in real life). Another potential problem is if the amount of traffic changing isn't continuous and so neither is the velocity. For example, consider a sudden accident. $\endgroup$ – memerson Nov 7 '18 at 1:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.