Answer
$R = 6.00~m~~$ and $~~\theta = 90.0^{\circ}$
Work Step by Step
It is given in the question that $r = 3.00~m$
The center of the circular path is $(0, 3.00~m)$
We can find the angle the particle has moved through at $t = 10.0~s$:
$\theta = (\frac{10.0~s}{20.0~s})(360^{\circ}) = 180^{\circ}$
We can find the x coordinate of the particle's position:
$x = (3.00~m)~cos~90^{\circ} = 0$
We can find the y coordinate of the particle's position:
$y = (3.00~m)+(3.00~m)~sin~90^{\circ} = 6.00~m$
We can find the magnitude of the particle's position vector:
$R = \sqrt{(0)^2+(6.00~m)^2} = 6.00~m$
The particle is directly above the origin. Therefore, the angle relative to the positive direction of the x axis is $90.0^{\circ}$
$R = 6.00~m~~$ and $~~\theta = 90.0^{\circ}$
