Is it true that among all the ways to travel a distance $X$ with an average speed $E$ in the speed vs. distance graph, traveling with a constant speed $E$ minimizes the time it takes to complete the travel? If yes, then is there any other way to travel for which it also takes that minimal time to complete the travel?

Note that I'm not talking about average speed (i.e. total distance divided by total time). What I mean by "an average speed $E$ in the speed vs. distance traveled so far graph" is that $E$ is equal to the average of the function $f: \text{distance} \to \text{speed}$, i.e. $f(x)$ is the speed of the object when the total distance it has traveled is $x$.


1 Answer 1


The answer is yes, traveling at constant speed is the unique way that minimizes travel time. I do hope there is an easier way to show this, but this is the best I can do at the moment.

The problem is the following: find the function $f(x)$ that minimizes $$T = \int\limits_0^X{\frac{dt}{dx}dx} = \int\limits_0^X{\frac{1}{f(x)}dx},$$ subject to the constraint $$EX=\int\limits_0^Xf(x)dx. $$

Let's transform this into an equivalent problem that is a bit simpler: define $$g(x) = f(x)-E.$$ Then, we must find the function $g(x)$ that minimizes $$T=\frac{1}{E}\int\limits_0^X{\frac{dx}{1+g(x)/E}}$$ subject to $$0=\int\limits_0^Xg(x)dx.$$

Note the equality: $$\frac{1}{1+z}=1 - z +\frac{z^2}{1+z},$$ you can verify this by simplifying the right hand side. Substituting this in the integral for $T$, $$T=\frac{1}{E}\left[\int\limits_0^X{dx} - \int\limits_0^X{\frac{g(x)}{E}}dx + \int\limits_0^X{\frac{g^2(x)/E^2}{1+g(x)/E}}dx \right].$$ The second integral is zero by the constraint equation, so $$T = \frac{X}{E}+\frac{1}{E}\int\limits_0^X{\frac{g^2(x)/E^2}{1+g(x)/E}}dx.$$ Note that as long as $g(x)>-E$ (you don't stop or change direction), the second term is always non-negative, and it is zero only when $g(x)=0$ (assuming it is a continuous function). Therefore, $g(x)=0$ (or $f(x)=E$) minimizes $T$ with the constraint that the "average" speed is $E$.

  • $\begingroup$ I edited my question to explain what I mean by average speed. $\endgroup$
    – avraham
    Jun 30, 2020 at 5:57
  • $\begingroup$ @avraham I will update my answer. Are you familiar with calculus of variations or the method of Lagrange multipliers? $\endgroup$
    – Puk
    Jun 30, 2020 at 6:15
  • $\begingroup$ I'm familiar with neither of them. $\endgroup$
    – avraham
    Jun 30, 2020 at 6:20
  • $\begingroup$ @avraham See the updated answer. $\endgroup$
    – Puk
    Jun 30, 2020 at 7:29
  • $\begingroup$ I understood everything apart from how you got to the first integral that is equal to T. $\endgroup$
    – avraham
    Jun 30, 2020 at 8:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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