Answer
$\vec{a_{avg}}=gsin\theta\approx3.3\frac{m}{s^2}$
Work Step by Step
We know that
$\vec{a_{avg}}=\frac{\vec v_f-\vec v_i}{\Delta t}$
We plug in the known values to obtain:
$\vec{a_{avg}}=\frac{10.0-0}{3.00}=3.3\frac{m}{s^2}$
Now we can compare it with $gsin\theta$
$\implies gsin\theta=(9.81\frac{m}{s^2})sin(20.0^{\circ})=3.3\frac{m}{s^2}$
Hence, we proved that the average acceleration of the skateboarder is $gsin\theta.$
