Answer
$\vec {a_{avg}}=(-0.15m/s^2)\hat x+(-3.73m/s^2)\hat y$
Work Step by Step
We find the components of $\vec {v_i}$ and $\vec{v_f}$
$\vec{v_i}=(4.10m/s)cos33.5^{\circ}\hat x+(4.10m/s)sin 33.5^{\circ}\hat y$
$\vec{v_i}=(3.42m/s)\hat x+(2.26m/s)\hat y$
The components of $\vec{v_f}$ are
$\vec{v_f}=(6.05m/s)cos59.0^{\circ}\hat x-(6.05m/s)sin 59.0^{\circ}\hat y$
$\vec{v_f}=(3.12m/s)\hat x+(-5.19m/s)\hat y$
Now $\Delta \vec v=\vec{v_f}-\vec{v_i}$
$\Delta \vec{v}=[(3.12m/s)\hat x+(-5.19m/s)\hat y]-[(3.42m/s)\hat x+(2.26m/s)\hat y]$
$\Delta \vec v=(-0.30m/s)\hat x+(-7.45m/s)\hat y$
We can find the average acceleration as
$\vec{a_{avg}}=\frac{\Delta \vec v}{\Delta t}$
$\vec {a_{avg}}=\frac{(-0.30m/s)\hat x+(-7.45m/s)\hat y}{2.00}$
$\vec {a_{avg}}=(-0.15m/s^2)\hat x+(-3.73m/s^2)\hat y$
