An ice-hockey puck slides east at 12 m/s…

Step 1. The change. $\Delta\vec{v} = \vec{v}_2 - \vec{v}_1 = (-12\hat{i} + 9\hat{j})\ \text{m/s}$.

Step 2. Its size and direction. $\sqrt{(12\ \text{m/s})^2 + (9\ \text{m/s})^2} = 15\ \text{m/s}$. It points west and north: $\tan^{-1}\dfrac{9\ \text{m/s}}{12\ \text{m/s}} = 36.9^\circ$ north of west.

Step 3. Average acceleration. $\dfrac{15\ \text{m/s}}{0.05\ \text{s}} = 300\ \text{m/s}^2$, in the same direction as $\Delta\vec{v}$.