Angular Position: t=0s, t=3.07s

[Nyx018]

This is the problem:
During a certain period of time, the angular position of a swinging door is described by θ = 4.92 + 10.7t + 1.91t2, where θ is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door at the following times.

then I'm given for part a) t=0s and part b) is t=3.07s

I know I have to use the equation for angular position s=r\theta then derive that once for angular velocity (\omega) then take the second derivative for angular acceleration (\alpha). I just don't understand what the values for s and r are and where the come from within this problem. i have that \theta(0)=4.92 or .0859rad

any help or guidance is much appreciated.

 

[heth]

s is the linear position.

Theta is the angular position (as described in the question).

That should get you started off.

 

[Nyx018]

ok I understand that but how do I find out or determine the value of the linear position from the formula theta?

 

[heth]

ok I understand that but how do I find out or determine the value of the linear position from the formula theta?

The part of the question quoted only asks for the angular position (another term for angle), angular speed and angular acceleration. There'd be no need to use s = rtheta at all, just differentiate what you've been given in the question.