Answer
$R = 5.54~m~~$ and $~~\theta = 67.5^{\circ}$
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 $t = 7.50~s$:
$\theta = (\frac{7.50~s}{20.0~s})(360^{\circ}) = 135^{\circ}$
We can find the x coordinate of the particle's position:
$x = (3.00~m)~cos~45 = 2.12~m$
We can find the y coordinate of the particle's position:
$y = (3.00~m)+(3.00~m)~sin~45 = 5.12~m$
We can find the magnitude of the particle's position vector:
$R = \sqrt{(2.12~m)^2+(5.12~m)^2} = 5.54~m$
We can find the angle relative to the positive direction of the x axis:
$tan~\theta = \frac{5.12~m}{2.12~m}$
$\theta = tan^{-1}~(\frac{5.12~m}{2.12~m})$
$\theta = 67.5^{\circ}$
$R = 5.54~m~~$ and $~~\theta = 67.5^{\circ}$
