Relativity of position and velocity in classical mechanics

["Don't panic!"]

I'm currently collating my own personal notes and would really appreciate some feedback on my description of the relativity of position and velocity in classical mechanics. Here is what I have written

"Position is clearly a relative quantity as two inertial frames S and S' displaced by a constant displacement vector \mathbb{r}_{0} will measure the position of an object to be at \mathbb{r} and \mathbb{r}' respectively, the two positions related by \mathbb{r} = \mathbb{r}' + \mathbb{r}_{0}. As these two frames are arbitrary and neither can be distinguished from the other as a preferred absolute rest frame (as a consequence of Galileo's principle of relativity), it must be that position is relative. This argument also holds if the two frames S and S' are in relative motion to one another, related by \mathbb{r} = \mathbb{r}'+\mathbb{v}t, where \mathbb{v} is the relative velocity between the two frames. Clearly it follows from this (by differentiating with respect to time) that velocity is also relative."

 

[Fredrik]

Sounds OK. You can also note that since v(t)=x'(t), and x(t) denotes the position coordinate in a particular coordinate system, it would be quite remarkable if the velocity function turns out to be the same in all coordinate systems. For this to be the case, the relationship between the position coordinate functions x and y associated with two different coordinate systems would have to be x(t)=y(t)+C, where C is a constant. This would imply that you can change from one of these coordinate systems to any other simply by doing a translation.

 

["Don't panic!"]

Isn't it true though that position is relative even if velocity isn't, i.e. if two frames are related by a constant translation $\mathbf{r}_{0}$ such that if $\mathbf{r}$ is the spatial position of an observed event in frame $S$ and $\mathbf{r}'$ is the spatial position of the same event in frame $S'$, then the positions they observe are related by $\mathbf{r} = \mathbf{r}_{0} + \mathbf{r}'$. As the displacement between them is constant we have that both observers $S$ and $S'$ will measure the same velocity for the observed event, as $\dot{\mathbf{r}} = \dot{\mathbf{r}'}$.