Measuring position, velocity and acceleration in relative motion

[Like Tony Stark]

Homework Statement

A particle moves with constant velocity $v_0$ in the $y$ direction with respect to an inertial system $A(x;y;z)$ as depicted in the picture. There is another system $B(x';y';z')$, which is not inertial and rotates with constant angular velocity $\omega$.
Determine
a) $r(t)$ from the perspective of $B$ and $A$
b) the velocity measured from $B$, i.e., the velocity relative to $B$
c) the acceleration measured from $B$, i.e., the acceleration relative to $B$

Relevant Equations

$\vec a=\vec a_B + \vec{\dot \omega} \times \vec r + \vec \omega \times (\vec \omega \times \vec r) + 2. (\vec \omega \times \vec v_{rel}) + \vec a_{rel}$

Well, $r(t)$ in $A$ is just a vector $(0;y)$ because is tangent to the trajectory. Then, from the perspective of $B$ the particle moves in an uniform circular motion. Is this right?

The velocity from $B$ must be $\omega$, right?

And what about acceleration?

 

[haruspex]

Well, r(t) in A is just a vector (0;y)

You need three dimensions, and it should be expressed in terms of v₀ and t.

from the perspective of B the particle moves in an uniform circular motion

No. That would be true if v₀=0.
If I read the diagram correctly, z'=z while the x' and y' axes rotate in A's xy plane.