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

Question

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.)

Answer 1

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.

Answer 2

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 $m<1<n$, 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 $1$.

Answer 3

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 $d_1(t_0 ) = 1$.

Answer 4

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.