Help with odd partial derivatives in velocity $\bar v^2 = \dot x ^2+\dot y^2$ I am doing a physics -course Tfy-0.2061. My teacher claims that this is velocity squared, $\bar v^2 = \dot x ^2+\dot y^2$. I cannot understand why it is not $\bar v^2 = (\dot x +\dot y)^2$. 
If distance is $\bar d = \bar x + \bar y$. Then velocity is $\partial_t \bar d = \dot x + \dot y$. Now just square it to get 
$$\bar v^2 = \dot x^2 +2\dot x\dot y +\dot y^2 \not = \dot x^2 +\dot y^2.$$
What does my teacher mean by velocity $\bar v^2 = \dot x ^2+\dot y^2$?
P.s. the goal was to do something called "nopeuden radiaalinen komponentti" that probably means radial component of velocity. I don't just understand what it means, some angular velocity? I am doing the exercise 3b here, sorry not in English.
Trial 1

The only way that my teacher can be correct is if $y_0=0$ and $x_0=0$ because



Distance is a vector quantity. Write it as $d=xi+yj$ where $i=(1,0)$, $j=(0,1)$. So $v=d'=x'i+y'j$. Now $v^2$ makes no sense - you can't square a vector. What makes sense is $\|v\|^2$ and that, the way vectors work, is $(x')^2+(y')^2$.
