Intuition behind why is unit speed parametrization and arc length parametrization the same?

Question

I have found a bunch of simple and not so simple proofs about why a vector function ($f(t)$) parametrized in such a way that it's derivative is always 1 ($|f'(t)|=1$) is the same as parametrizing it by arc length ($f(s) \iff |f'(s)|=1$).

Just to provide one example of a proof found, I write down the definition for arclength with $t$ substituted with $s$ $$s=\int_0^s{|f(s)|}ds$$ apply derivative by both sides with respect to $s$ $$1=|f(s)|$$ BOOM!, proven. (as long as the reparametrization is a biyective, smooth and has an inverse)

The question is, How can i understand this as an intuitive thing? I think im missing the "aha" moment where is makes sense that an arc length function would have unit speed.

Answer 1

Since the question was about intuition ...

If you walk at a constant speed of $1$, then the distance you've covered in time $T$ is $T$.

Answer 2

I think the intuitive argument that I would use here is to think of the curve $f$ as taking a real interval $(a,b)$ to some curve segment $f((a,b))$ in a manifold or Euclidean space. Having $||f'(s)|| = 1$ for all $s$ is kind of like saying that "$f$ traces out the curve segment just as quickly as it passes through values in $(a,b)$".

Imagine if $||f'(s)|| >1$, then the curve segment would get traced out quicker than the interval $(a,b)$ meaning that the curve segment would have a length shorter than $b-a$. Similarly, if $||f'(s)|| < 1$ it would take a longer interval $(a,b+\epsilon)$ to trace out the curve hence the segment would have a length longer than $b-a$.