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?)

  • $\begingroup$ 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. $\endgroup$ – senreigh Aug 8 '19 at 3:21
  • $\begingroup$ @senreigh but they said infinitesimal precision. I take that to mean integrate. $\endgroup$ – user658409 Aug 8 '19 at 3:21
  • $\begingroup$ 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 $\int_{20}^{60} \frac{1}{60-x}dx$. $\endgroup$ – senreigh Aug 8 '19 at 3:28
  • 1
    $\begingroup$ Can OP clarify whether they mean discrete or continuous? $\endgroup$ – user658409 Aug 8 '19 at 3:29
  • $\begingroup$ 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. $\endgroup$ – 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 $\lim_{t\to\infty} x(t)=60$.


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