# If Zeno could drive: Traveling at a speed (in miles per hour) that always exactly matches the distance (in miles) to the destination

As I was driving on the highway this afternoon, I thought to myself: what if, at each moment, I were to move at a speed that matched exactly the distance I had remaining? As an example, at 60 miles from the destination I would drive at 60 miles per hour; then, with 59 miles remaining, I would slow to 59 miles per hour; and so on, of course with infinitesimal precision. Two questions emerge from this situation:

1. How long will it take to travel from mile-marker 60 to mile-marker 0?
2. And how long will it take to travel from mile-marker 60 to, say, mile-marker 20?

I've read through several previous postings of this same question (here and here, for example) but have yet to find a satisfying, thorough explanation. (How exactly, for example, do we involve the harmonic number $$H_{60}$$ to this problem?)

• This isn't an infinite series it is just a sum $\sum_{i=0}^n \frac{1}{60-i}$ where $n=60-m$ where m is the mile marker you stop at. – senreigh Aug 8 '19 at 3:21
• @senreigh but they said infinitesimal precision. I take that to mean integrate. – user658409 Aug 8 '19 at 3:21
• This is discrete though once you get to 1 you go 1 mile per hour for 1 hour then stop so it doesn't involve an integral, if you were continuously slowing down however then the integral $\int_0^{60} \frac{1}{60-x}dx$ diverges. but if you stop at say 20 then its fine $\int_{20}^{60} \frac{1}{60-x}dx$. – senreigh Aug 8 '19 at 3:28
• Can OP clarify whether they mean discrete or continuous? – user658409 Aug 8 '19 at 3:29
• I think both are interesting cases to consider, actually, so I appreciate both of the solutions posted here. Initially, though, I was taking deceleration to be continuous. – nbogs Aug 11 '19 at 2:24

Your velocity is $$v(t)=60-x(t)$$ where $$x(t)$$ is the distance from start. Note that $$v(t)=x'(t)$$ so this is $$x'(t)=60-x(t)$$. Solution to this differential equation is $$x(t)=60+Ce^{-t}$$. Note that $$x(0)=0$$ and so $$x(t)=60-60e^{-t}$$. Note that for all finite time $$t$$ we have $$x(t)<60$$ (i.e. you aren't at your desination). However $$\lim_{t\to\infty} x(t)=60$$.