Explanation

Average acceleration = change in velocity/time interval = $\Delta \vec{v}/t$ $\vec{v}_{1} = 5\hat{i}, \vec{v}_{2} = 5\hat{j}$ $\Delta \vec{v} = (\vec{v}_{2} - \vec{v}_{1})$ $= \sqrt{(v_{1}^{2} + v_{2}^{2} + 2v_{1}v_{2}\cos 90)}$ $= \sqrt{(5^{2} + 5^{2} + 0)}$ $[ As|v_{1}| = |v_{2}| = 5 m/s]$ $= 5\sqrt{2} m/s$ $Avg. acc. = \Delta \vec{v}/t = 5\sqrt{2}/10 = 1/\sqrt{2} m/s^{2}$ $\Rightarrow \tan \theta = 5/ - 5 = - 1$ which means $\theta$ is in the second quadrant. (towards northwest)