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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.

wannabemathman
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    "It seems that average velocity over some time interval....will be equal to the instantaneous velocity at the midpoint ..."

    Why at the midpoint? There's no definition of the position vs time, so one can't make this argument.

    Are you studying the mean value theorem?

     – MaximusFastidiousIrreverence Aug 05 '22 at 19:09
  • You are looking for the point where the slope of the tangent line is equal to the slope of the secant line. – John Douma Aug 05 '22 at 19:22
  • To answer your question Maximus, I'm studying physics 1 at the moment, and had not thought about the mean value theorem - I appreciate you turning me in that direction. as I understand it, that applies generally, but this is at least a slightly more specific case. true there's no definition of the position vs time graph, except that the second derivative wrt time is constant. so I'm wondering if that's enough of a constraint to make the above statement true. – wannabemathman Aug 05 '22 at 20:14

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

Mark Saving
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  • this is amazing - thank you! because if the acceleration is constant, the average of two velocities is equal to the formal "average velocity" over that interval, correct? that makes sense to me intuitively, although I do struggle a little bit with the idea behind the average of two velocities being the same as the average velocity. there must be a mathematically rigorous explanation behind that idea - I'll need to do some digging around I think. – wannabemathman Aug 05 '22 at 20:20
  • @wannabemathman The fact that the average velocity is the velocity at the midpoint follows from the fact that velocity is a linear (in the high school sense) function of time. It’s easy to see this geometrically. Given two points on a line, the height of their midpoint is the average of the two heights. – Mark Saving Aug 05 '22 at 23:58
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Alternative approach:

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

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

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

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

user2661923
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  • very cool! my book isn't using integrals (yet), but if I'm understanding you correctly, the first line is the average velocity? so the integral in the numerator is the change in position? – wannabemathman Aug 05 '22 at 20:40
  • @wannabemathman Close. Within the integral in the numerator, the velocity is expressed as a function of time. Intuitively, suppose that you graph a function $~f:\Bbb{R} \to \Bbb{R}.~$ Then, the value of the function at any point $(x = x_0)$, expressed as $f(x_0)$, shows as the height of the graph at $(x = x_0).$ Then, the following expression can be thought of as the area under the graph, from $(x=a)$ to $(x=b)$: $$\int_a^b f(x) dx.$$ Then, if you divide this area by the width of the interval, $(b-a)$, you get the average height of the function, in this interval. – user2661923 Aug 05 '22 at 22:14