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.