I forgot to tell about $\theta$ as in the editted question, so $vt$ can be replaced with $\theta$ right?

No wait, how did you get Vt/R? I don't understand.

If the (constant) speed is $v$, and $R$ is the radius of the circle, then the angular speed is $v/R$.

To see this, suppose the angular speed (the coefficient of $t$ inside the $\cos$ and $\sin$ functions) is $\omega$, so $\vec{Q} = R\cos(\omega t)\hat{i} + R\sin(\omega t)\hat{j}$. Then as $t$ goes from $0$ to $\frac{2\pi}{\omega}$, the particle $Q$ describes exactly one full circle - so the total distance moved is the circumference $2\pi R$. Then the (constant) speed is $\text{speed} = \dfrac{\text{distance}}{\text{time}} \Rightarrow v = \dfrac{2\pi R}{2\pi/\omega} = R\omega \Rightarrow \omega = \dfrac{v}{R}$. A shorter way is to differentiate $\vec{Q}$ and equate the magnitude to $v$.

Oh yeah what was I thinking. Thanks.

@S.Dan I'll work in that $\theta$ and update the answer when I'm done.

FINALLY, was able to log in. There was some issue with my browser or something, and I was not able to sign into chat

Ah, that does help, thanks! Lemme work it out.

So $r = R\tan 30 = R/\sqrt{3}$, right?

Yep

173.2

The $x$ and $y$ components of $PQ$ are $R\cos \omega t$ and $R\sin \omega t + r = R\sin \omega t + R\tan 30$. So $\tan \theta = (R\sin \omega t + R\tan 30)/(R\cos \omega t) = \tan \omega t + (\sec omega t)(\tan 30)$

Where $\omega = v/R$, of course.

Maybe you should write this down. The LateX isn't rendered here, right?

But how can I get rid of the 't'?

You can't, because it depends on time. See how in the question it tells you to determine the angular velocity and acceleration.

If the velocity were independent of t, the acceleration would be 0

The angular velocity of an object moving with uniform speed around a circle is constant only when measured from the center of the circle (or a point directly above or below it in three dimensional space). Measured from any other point, the angular velocity will vary with time (because the distance from that point will also vary with time).

Oh I see. Great work (y). So what should be done to t?

It's part of the answer. Non-uniform quantities have to have t as part of their equation.

Hmmn.. but there i no relevence of 't' in da question :/ anyway,I'll figure it out. Thanks alot :)