Average velocity is usually expressed as the displacement divided by the time. For example,
V_avg= (x(t)-x_o)/t .
I $$v_{\text{avg}} = \frac{x(t)-x_0}{t}$$ I have not, however seen it expressed in terms of acceleration. Would it be incorrect to say $$v_{\text{avg}}= \bigl(a(t)-a_0\bigr)\times t$$
V_avg= (a(t)-a_0)*t.