Average velocity is usually expressed as the displacement divided by the time. For example, $$v_{\text{avg}} = \frac{x(t)-x_0}{t}$$ I have not, however seen it expressed in terms of acceleration. Would it be incorrect to say $$v_{\text{avg}}= \bigl(a(t)-a_0\bigr)\times t$$
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$\begingroup$ The definition of the average of a function is $\int_a^bf(x)dx/\int_a^bdx$. Can you see why the top equation corresponds to this definition but the bottom one doesn't? $\endgroup$– Johnathan GrossCommented Sep 6, 2018 at 1:52
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Take for example $x(t)=10\,\text{m}$, $x_0=0\,\text{m}$, $a(t)=6\, \frac{\text{m}}{\text{s}^2} $, $a_0=1 \,\frac{\text{m}}{\text{s}^2} $ and $t=5\text{s}$. Then the correct formula gives $$v_{\text{avg}}=2\, \frac{\text{m}}{\text{s}} $$ whereas your formula gives $$v_{\text{avg}}=1\, \frac{\text{m}}{\text{s}} .$$
answered Sep 6, 2018 at 1:44