Curves with "constant speed"? I am new to the concept of curves.
Let us a assume we have a simple function such as $f:\mathbb{R}^+\rightarrow \mathbb R^+\quad f(x) := \sqrt{x}$. (Or $f(x)=\exp(x)$ or a polynomial etc.).
We can turn it into a curve $F: [a,b] \rightarrow \mathbb R^+ \times \mathbb R^+$ with $F(t):= (x(t), f(x(t)))$, right?
Obviously, we could set $x(t)=t$. But, how could we achieve that the  curve has a "constant speed", thus arc-length is proportional to the time interval:
$$\exists_{c>0}\forall_{t_1\in [a,b]}\forall_{t_2\in [a,b]}\quad\text{arc_length}(F(t_1),F(t_2)) = c |t_1-t_2|$$
Also how is this property typically called?
Edit: 
Let me rephrase my question, to clarify what I am looking for. Assume a turkey is walking on a curve (e.g. log(x)) with constant speed. How can I calculate it's position given the time passed? If there is no closed-form solution, any numerical method (for a general function $f$) would be great too. 
 A: For a parametric curve $(x(t),y(t))$ we need that for arc length $s$ it should satisfy $$\frac{ds}{dt} = \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2} = \operatorname{const.}\tag1$$
So for the curve $(t,f(t))$ this gives
$$\sqrt{1+\left(\frac{df}{dt}\right)^2} = \operatorname{const.}$$
which implies that $f^\prime(t) = \operatorname{const}$ and thus $f(t) = mt+c$ is linear. Linearity of $f(t)$ arises from simply interpreting a graph as a curve. There are other solutions for which the property holds, e.g. $(a\cos(\omega t), a\sin(\omega t))$ which gives
$$\frac{ds}{dt} = \sqrt{\left(- a\omega \sin(\omega t)\right)^2+\left(a\omega \cos(\omega t)\right)^2},$$
$$= \sqrt{(a\omega)^2(\sin^2(\omega t)+\cos^2(\omega t))},$$
$$=a\omega.$$
Note that this means that up to some constant factor the curve is also parameterized by arc length! As for the last part of your question, I'm not aware of what this property is called.
Answer to edited question:
At a given time we have moved $vt=s$ along the curve, but we know from equation (1) that $ds/dt = v$ which gives
$$s = vt = \int_0^t \sqrt{\left(\frac{dx}{d\tau}\right)^2 + \left(\frac{dy}{d\tau}\right)^2} d\tau$$
Since $y = f(x)$ we can use the chain rule
$$vt = \int_0^t \sqrt{\left(\frac{dx}{d\tau}\right)^2 + \left(\frac{df}{dx}\frac{dx}{d\tau}\right)^2} d\tau$$
$$ = \int_0^t \frac{dx}{d\tau} \sqrt{1 + f^\prime(x)^2} d\tau$$
$$ = \sqrt{1 + f^\prime(x)^2} \int_0^t \frac{dx}{d\tau} d\tau$$
$$\Rightarrow vt = \sqrt{1 + f^\prime(x)^2}\,\cdot\, x \tag2$$
where we have assumed that $x(0)=0$ (we can always shift our time axes so that this is the case).
If $f^\prime$ is invertible and simple enough then equation (2) is exactly solvable. Once we know $x(t)$ we then can simply plug this into $f$ to get $f(x(t))$. For instance if our curve is linear $f(x) = mx+c$ we get
$$x = \frac{vt}{\sqrt{1+m^2}},$$
if $m=0$ this gives us $x=vt$ as expected, similarly for $m=1$ we get $x=vt/\sqrt{2}$ as expected.
In general however this equation is not solvable (I think) thus you could use some numerical method (for example Newton-Raphson) to find the roots of the equation
$$(vt)^2 - x^2(1 + f^\prime(x)^2)=0.$$ 
A: You may achieve an arc-length parametrization theoretically; practically you won't get such in most cases.  That's the bad news.  Now for the good news: practically and theoretically one doesn't need an explicit arc-length parametrization in most case but the derivative of a curve in respect to the arc length.  For any immersive $C^1$ curve $c$ it's derivative in respect to arc length is given by
$$\frac{\dot c}{\|\dot c\|}.$$
