# Consider a man who travelled exactly 2 km in two hours. Is there a one-hour interval when he traveled exactly 1 km?

Question :

Consider a man who travelled exactly 2 km in two hours.
Is there a one-hour interval when he traveled exactly 1 km?

Can we make a mathematical argument?

I have written my attempt in an answer below. Does anyone else have a better approach?
Are my assumptions not necessary, or we can produce counter examples without them? Can we make-do with some intermediate assumptions?

(This is not the same as the Universal Chord question because there's no $$f$$ with $$f(0)=f(1)$$ and there's no continuity assumption. Answers to the cyclist question are not satisfactory.)

• No continuity assumption? Can we suppose he was simply teleported to the destination at some point in time? Commented Jan 14 at 10:28
• @whoisit, the distance function $d$ does have to be continuous. (What I think you mean is that it is not necessarily differentiable.) So the Universal Chord Theorem applies here to the function $f(t)=d(t)-t$. Commented Jan 14 at 10:57
• Interestingly, you need that the total distance is an integer for this to be true. For example, cover 1.5km in 1.5 hours by covering 0.5km segments steadily in 20 minutes, then 50 minutes, then 20 minutes. Every 1km segment is traversed in exactly 70 minutes, although the average pace is 1km/hr. (Note every 1km segment contains the middle 0.5km plus a total of 0.5km from the first and last portions.) Discussed here: arxiv.org/abs/1507.00871
– usul
Commented Jan 15 at 5:27
• While the upvoted answer is certainly correct, the short answer already appears in the comments: this is exactly the Universal Chord Theorem, applied to $f(t)=d(2t)-2t$. You have $f(0)=f(1)=1$ and, yes, $f$ is continuous. You can conclude that there are time intervals of length exactly $1$ hour, $40$ minutes, $30$ minutes, $24$ minutes, $20$ minutes, etc., over which the runner’s average velocity is $1$ km/hr. Commented Jan 15 at 16:40
• @Mazura: I believe the question is whether, for all travels that meet the criteria described in the question, there must be such an interval; your comment merely points out that there is some travel that has such an interval (and, as the answers show, we can actually say much even if the velocity was not constant). Commented Jan 15 at 19:31

Let $$d(t)$$ be the distance travelled at time $$t$$ (in hours). We have $$d(0) = 0$$ and $$d(2) = 2$$. It is safe to assume that $$d$$ is continuous. We want to find a $$t\in[0;1]$$ with $$d(t+1) - d(t) = 1.$$ Define an auxiliary function $$f(t) = d(t+1) - d(t) - 1.$$ Thus, our condition becomes $$f(t) = 0$$.

We have that \begin{align*} f(0) &= d(1) - d(0) - 1 = d(1) - 1 \\ f(1) &= d(2) - d(1) - 1 = 1 - d(1) = -f(0) \end{align*} Thus:

If $$f(0) = f(1) = 0$$, then we are done.

If $$f(0) \neq 0$$, we have that $$f(0)$$ and $$f(1)$$ have different signs. By the intermediate value theorem, there is a $$t\in (0;1)$$ with $$f(t) = 0$$, as desired.

Let's define $$d$$ without any velocity assumption: Let $$p(t)$$ be the position of the man - and define $$d(t)=p(t+1)-p(t)$$. As teleportation is not allowed, $$p$$ is continuous, thus so is $$d$$.

We want to find whether there is always a $$t$$ where $$d(t)$$ takes the value 1 km. We know that $$d(0)+d(1)=2$$.

WLOG assume $$m = d(0) \leq d(1) = n$$. Then, either $$m=n=1$$ in which case we're done, or we have $$m<1, in which case we can use continuity of $$d(t)$$ to invoke Intermediate Value Theorem to claim that there was some $$t$$ in between where the value was $$1$$.

• You don't need any assumptions about the velocity, as Tobius' answer shows. Commented Jan 14 at 10:56
• I find this answer much more intuitive than Tobius'. However, the bit about integrals is superfluous. You can define $d(t)$ as "distance traveled between $t$ and $t+1$" and everything in your answer remains correct, and this correctly proves the existence of $t$ such that $d(t) = 1$. You don't need to know that $d(t) = \int_{t}^{t+1}v(t) dt$ and so you don't need any assumptions about integrability or continuity of the velocity (or even existence of a velocity function at all). All you need to know is that $d(t)$ is continuous, which is equivalent to "the traveler did not teleport".
– Stef
Commented Jan 14 at 20:04
• "or has discontinuities at a finite number of points" - and the question has no meaningful answer? - At 2000mph that takes 4 seconds. So between 0:0:02 and 1:0:02 they went one mile. That's assuming they didn't just stand there between 0 seconds and 1:59:56 hours. Commented Jan 15 at 19:49

With the continuity assumption, define a function on the interval $$t \in [1,2]$$ by $$d_1 (t) \equiv d(t) - d(t-1)$$, the distance covered within the last hour, where $$d(t)$$ is the position at time $$t$$. Then $$d_1 (1) = d(1) - d(0) = d(1)$$, and $$d_1 (2) = d(2) - d(1) = 2 - d(1)$$. The mean of the two boundary values $$d(1)$$ and $$2 - d(1)$$ of the function $$d_1 (t)$$ is $$1$$, therefore $$1 \in [d_1 (1), d_1 (2) ]$$ no matter what $$d(1)$$ is, and by the intermediate value theorem there exists a $$t_0 \in [1,2]$$ such that $$d_1(t_0 ) = 1$$.

Here is a very Simple Solution :

## SUMMARY :

When Initial Window + last window = 2km, either both widows are 1km or one is less & the other is more , in which case , there must be some Intermediate Window where the continuous window Distance $$D(t)$$ changes over , via becoming equal to 1km.

## DETAILS :

Let us take a sliding window of 1 hour duration : $$(t,t+1)$$
That window starts at $$t=0$$ ( window is $$(0,1)$$ ) & ends at $$t=1$$ ( window is $$(1,2)$$ )

Let Initial window have Distance $$D(t)$$ less than 1km.
When window moves forward by small $$h$$ then new Distance ( within that Window ) will vary to $$D(t+h)$$
Due to continuity , Distance might either become exactly 1km (& we are DONE) or be still less than 1km.

Eventually , we will reach the last window $$(1,2)$$.
When that window still has less than 1km , then we see that total Distance traveled is less than 2 km (Initial window + last window)
Hence that last window must be 1km or more. Else there is a Contradiction.
Hence , we must have some Intermediate Window where "$$<1km$$" change to "$$>1km$$" , or rather that Intermediate Window must have "$$=1km$$".
DONE

We have similar argument when Initial Window is more than 1km , where some Sliding Window must have 1km. When all Intermediate Windows are still more than 1km , then the last window must have 1km. When last window still has more than 1 km , total Distance is more than 2km which is a Contradiction.
We are DONE with all Cases.