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First, I wanna clarify that I understand the intuition behind the definition of velocity in 1, 2 and 3-D (The derivative of the position vector).

But the other day I was thinking if that definition agrees with the intuition behind constant speed (just in modulus). I mean, when somebody says "X was moving with constant speed" one automatically thinks that X is moving "softly", or, more precisely, it goes through a certain distance always in the same time(that's why speed is measured in lenght/time, I guess). That distance is not measured with the displacement vector, as far as I see. It is the length of the path that the particle has described. So, I believe that there should be a formal proof, that if $r(t)$ is function from $\mathbb{R }$ to $\mathbb{R}^2$ (or to $\mathbb{R}^3$), and $r'(t)=v$, and $|v|$ is constant, then the function $r(t)$ describes a curve that has the same lenght in each interval of, say, lenght $[0,1]$ (1 second maybe). Does such proof exist?

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  • $\begingroup$ The proof is trivial. The distance covered in unit time interval by a body with constant speed is $$\text{distance=} \int_{t_o}^{t_o +1} |v| dt= |v|$$. $\endgroup$ – Omar Nagib Aug 2 '16 at 5:57
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The distance traveled in time $t$ along a path $\gamma$ is by definition the integral over the speed

$$ L = \int_0^t \lvert{\dot\gamma(s)}\rvert ds $$

which for constant speed $v$ becomes $L(\gamma) = v \int_0^t ds = vt $

This is linear in $t$ and hence proofs your conjecture. This is no wonder, because speed is defined as distance over time.

Velocity on the other hand is a different thing. On a circular path for example, velocity changes constantly while speed may very well be constant.

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