Answer
$v_{ave} = 0.85~m/s~~$ 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 $(0, 3.00~m)$
In part (d), we found that $~~R = 4.24~m~~$ and $~~\theta = 135^{\circ}$
We can find the magnitude of the average velocity:
$v_{ave} = \frac{4.24~m}{5.00~s} = 0.85~m/s$
Note that the direction of the average velocity is always in the same direction as the displacement.
$v_{ave} = 0.85~m/s~~$ and $~~\theta = 135^{\circ}$
