Answer
$(-9.8\frac{m}{s^2})\hat y$
Work Step by Step
We can find the average acceleration as follows:
$\vec{a_{avg}}=\frac{\vec {v_f}-\vec{v_i}}{\Delta t}$
We plug in the known values to obtain:
$\vec{a_{avg}}=\frac{(-4.5\frac{m}{s})\hat y-(4.5\frac{m}{s})\hat y}{0.92}$
$\vec{a_{avg}}=(-9.8\frac{m}{s^2})\hat y$
