Answer
$\theta=45^{o}\quad$ ($45^{o }$ north of due east)
Work Step by Step
In part (e), we found
$ \vec{a}_{avg}=(0.333m/s^{2})\hat{ i}+(0.333m/s^{2})\hat{j}$
This vector points to quadrant I.
The angle is
$\displaystyle \theta=\tan^{-1}\frac{a_{y}}{a_{x}}=\tan(1)=45^{o}$
$\theta=45^{o}\quad$ ($45^{o }$ north of due east)
