Answer
$R = 4.24~m~~$ and $~~\theta = 135^{\circ}$
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 $t = 5.00~s$:
$\theta = (\frac{5.00~s}{20.0~s})(360^{\circ}) = 90.0^{\circ}$
Then the particle is at the position $(3.00~m, 3.00~m)$
We can find the angle the particle has moved through at $t = 10.0~s$:
$\theta = (\frac{10.0~s}{20.0~s})(360^{\circ}) = 180^{\circ}$
Then the particle is at the position $(0, 6.00~m)$
We can find the components of the displacement:
$\Delta x = 0~m-3.00~m = -3.00~m$
$\Delta y = 6.00~m-3.00~m = 3.00~m$
We can find the magnitude of the displacement:
$R = \sqrt{(-3.00~m)^2+(3.00~m)^2} = 4.24~m$
We can find the angle above the negative direction of the x axis:
$tan~\theta = \frac{3.00~m}{3.00~m}$
$\theta = tan^{-1}~(\frac{3.00~m}{3.00~m})$
$\theta = 45.0^{\circ}$
Therefore, the angle relative to the positive direction of the x axis is $~~135^{\circ}$
$R = 4.24~m~~$ and $~~\theta = 135^{\circ}$
