# Is it possible for two cars to leave an origin at different times and arrive at the destination at the same time?

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?

• "starts at X, goes up to Y" That's beyond vague. Also, what efforts have you tried to put into solving the problem yourself? – Don Thousand Nov 7 '18 at 1:08
• 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 – Y L Nov 7 '18 at 1:11
• 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. – Don Thousand Nov 7 '18 at 1:13
• Rushabh, do you disagree with the answer below? You said it IS possible. – Y L Nov 7 '18 at 1:45
• I do agree it is possible. I just think the question is quite poor. – Don Thousand Nov 7 '18 at 2:32

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.