Assume constant acceleration. It seems that average velocity over some time interval [t1, t2], will be equal to the instantaneous velocity at the midpoint t = 1/2[t1 + t2]. I'm wondering how you might prove this mathematically (assuming what I've said is even true). If it's not always true, I would be curious to see a counterexample. Thank you kindly, in advance!
I suppose it's really a question about secant lines, tangent lines, and derivatives, but this is the context in which I had the thought. Thanks again for taking the time.
Best Answer
This follows immediately from the definition of constant acceleration. We can define constant acceleration to mean that there is some acceleration $a$ such that for all times $w, x$, we have $v(w) - v(x) = a(w - x)$.
Suppose the constant acceleration is $a$. Let $m = \frac{t_1 + t_2}{2}$.
Then $v(m) - v(t_1) = a (m - t_1)$. Similarly, $v(t_2) - v(m) = a (t_2 - m)$.
Note that $m - t_1 = t_2 - m$. Therefore, $v(t_2) - v(m) = v(m) - v(t_1)$. Therefore, $v(m) = \frac{v(t_1) + v(t_2)}{2}$.