Prove when Instaneous Velocity is equal to Average Velocity with Constant Acceleration

Question

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.

Answer 1

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, $v(m) = \frac{v(t_1) + v(t_2)}{2}$.

Answer 2

Alternative approach:

$$\frac{\int_{t_1}^{t_2} \left[V_0 + at\right]dt}{t_2 - t_1}$$

$$= \frac{\left[V_0t + \frac{1}{2}at^2\right] ~|_{t=t_1}^{t=t_2}}{t_2 - t_1}$$

$$= \frac{\left[V_0(t_2 - t_1)\right] + \left[\frac{1}{2}a(t_2 - t_1)(t_2 + t_1)\right]}{t_2 - t_1}$$

$$= V_0 + \left[\frac{1}{2}a(t_2 + t_1)\right].$$