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During a move, Driver A and Driver B drove two separate cars from their home town, Washington D.C. to New York City. It took them about 5 hours to travel to NYC. They agreed to travel no faster than 80mph during the trip. The total distance between the two places is 240 miles. Using the Intermediate Value Theorem, show that they must drive at the same speed at one time or another during the entire journey.

I am not completely sure what I should be doing for this question. Typically I see a given function within the questions, and a request to find the root or a given point within the function. I do know that velocity is a continuous function, so IVT applies. I also know that regardless of the speeds traveled, the speed of the slower of the two cars will be achieved by the faster car at least at the tail-end of the trip during deceleration. I am confused as to how to portray this information, if it is even the correct information to portray. Thank you in advance.

insipidintegrator
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gh0st
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    Hint: For rigor, use IVT on $v_1-v_2$ where $v_1,v_2$ are the speeds. Show that $v_1-v_2$ Must be zero at some point of time. – insipidintegrator Jul 09 '22 at 05:24
  • Thank you for the hint. That is as much as I figured, since the difference must be 0 at some time. Would a random speed between 0 and 80 be chosen by me and using that to prove? Or is it about how I phrase the answer? – gh0st Jul 09 '22 at 05:30
  • Did the drivers travel the same distance in same time? Like, both covered 240 miles in 5 h? IF YES, then try to prove a contradiction occurs if, say, driver A always travels faster than B, and also if A always travels slower than B. Thus A must be slower than B at some point and faster than B at some other point. So the function of the difference in speeds is negative and positive at different times. – insipidintegrator Jul 09 '22 at 05:50
  • I would assume that since it says, "It took them about 5 hours..." that they made the trip in the same amount of time. – gh0st Jul 09 '22 at 06:40
  • Would this be on the right track?

    Assuming that one driver leaves before the other, V(a) is a function which describes the relative velocity of driver a during the trip. Let v1 represent the velocity of driver a, v2 represent the velocity of driver b, and N represent the difference between the drivers. If driver a leaves first: at the beginning of the trip V(a)=v1-v2=+N and at the end of the trip V(a)=v1-v2=-N. Since V(a) is a continuous function and runs from -N to +N, N must be zero at some time, and both drivers were at the same speed at least once during their trip.

     – gh0st Jul 09 '22 at 07:59
  • Wait. So they start at different times? I assumed they start together. – insipidintegrator Jul 09 '22 at 08:09
  • There is no specification on when they start or stop, just that they both traveled about 5 hours. It also doesn't say one traveled faster than the other. I meant "earlier" as in even a couple of minutes before. – gh0st Jul 09 '22 at 08:10
  • Ok. Let me post an answer and then you can ask for clarifications. – insipidintegrator Jul 09 '22 at 08:16

1 Answers1

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I think that the relative starting times does not matter.

Let car A start at $t=t_1$ and B start at $t=t_2$. We assume that both cars take 5h to travel $240$ miles. Just for the sake of it, average speed is 48 mph. Let $v_1(\Delta t)$ and $v_2(\Delta t)$ be instantaneous speeds of A and B at $t=t_1+\Delta t$ and $t=t_2+\Delta t$ respectively, where $0\leq \Delta t\leq 5h$. Since in a real-world scenario, speed must be continuous, hence $v_1,v_2$ and thus $v_1-v_2$ are continuous functions.

Consider $v_1(\Delta t)\gt v_2(\Delta t)$ for all $\Delta t.$ This implies that $$\int_0^{5h}v_1d(\Delta t)>\int_0^{5h}v_2d(\Delta t)$$ which implies that the distance travelled by A in 5 hours is greater than that by B in 5 hours, which is clearly not the case as they are equal. You can similarly do the vice-versa.

This implies that there must exist at least one value of $\Delta t$ such that $v_1(\Delta t)>v_2(\Delta t)$ AS WELL AS at least one value of $\Delta t$ such that $v_1(\Delta t)\lt v_2(\Delta t)$.

Thus, the function $v_1-v_2>0$ for some $\Delta t$ as well as $v_1-v_2<0$ for some other $\Delta t$ on the interval $\Delta t\in [0,5hrs]$. Thus, by the intermediate value theorem, $v_1-v_2$ must have a zero in $\Delta t\in[0,5h]$. Hence proved.

The point is not that they will have the same velocity at one instant of time. The point is that they will have the same velocity corresponding to the same time interval $\Delta t$ after they start.
This will change to having same speeds at same instants of time if they start together.

insipidintegrator
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  • This makes sense and I believe I understand more about this on multiple levels. Many thanks! – gh0st Jul 09 '22 at 08:45
  • You’re welcome. – insipidintegrator Jul 09 '22 at 08:46
  • However, I would still like to know whether the “80mph” data is simply superficial or actually useful. I tried to analyse it with simple velocity-time graphs but realised quickly that I was just zoning on extremely specific cases. So if you find that information anytime, please comment on my answer to let me know. Thanks! – insipidintegrator Jul 09 '22 at 08:47
  • I have been stuck attempting to do it with the 80mph, which I think was where my thoughts kept faltering the same, hyper-focusing on velocity graphs. I'll have to find out if it is useful data or just random data and come back with that info later. – gh0st Jul 09 '22 at 09:13