The first result in physics texts is the distance travelled under constant acceleration $a$ and initial velocity $v_0$ is given by $$x=x_0+v_0t+\frac{1}{2}at^2$$ This can be proved by calculating the area under the velocity curve, either by using calculus or geometrically.

However, a proof in several texts namely, Giancoli, and that of Jearl Walker is too claim that under constant acceleration, the average velocity $\overline{v}(=\frac{x-x_0}{t})$ satisfies, $$\overline{v}=\frac{v_0+v}{2}$$ Indeed this is true and it implies the above formula, but I do not see any proof of this equation that does not assume the conclusion. The books seem to emphasize the word "average", but this is not a proof. A proof is a sequence of equations. Am I right about this ? Or is there indeed a mathematical proof this equation? What do people think about this?

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    $\begingroup$ Compute the time-weighted average of velocities. See also a Plot of v vs t for constant acceleration. Find the constant velocity that has the same displacement in the same elapsed time. $\endgroup$
    – robphy
    Nov 26, 2022 at 0:04

2 Answers 2


Too easy.

The average velocity at time t is just the distance it traveled at time t divided by the time.

$$\bar{v} = D(t)/t = \frac{x(t) - x_0}{t} = v_0 + \frac{1}{2}at$$

But $v(t) = v_0 + at $ is just the velocity at time t. So that gives this. $$at = v(t) - v_0$$

And putting that back in the $\bar{v}$ equation gives you this.

$$\bar{v} = v_0 + \frac{1}{2}(v(t)-v_0) = \frac{1}{2}(v(t)+v_0) $$

QED, no integrals required.


Sadly, "average velocity" is rarely defined explicitly as "the time-weighted average of velocities". From this, one can then express this as the "total displacement over the elapsed time".

Thus, I write $\displaystyle\bar v \equiv\frac{ \int v\ dt }{\int dt}$, not $\displaystyle \bar v \equiv\frac{\Delta x}{\Delta t}$.

\begin{align} \bar v &\equiv\frac{ \int v dt }{\int dt} =\frac{\Delta x}{\Delta t}\\ &\stackrel{const\ a}{=}\frac{ \int (v_0+at) dt }{\int dt}\\ &\stackrel{const\ a}{=}\frac{ v_0t + \frac{1}{2}at^2 }{t}\qquad =\frac{\Delta x}{\Delta t}\\ &\stackrel{const\ a}{=}\frac{ v_0t + \frac{1}{2}(\frac{v-v_0}{t})t^2 }{t}\\ &\stackrel{const\ a}{=}\frac{ v_0+v}{2}\\ \end{align} For the case of constant acceleration, this happens to be the arithmetic sum of the initial and final velocities.

If you plot $v$-vs-$t$ for constant acceleration, you get a line.
Compute the time-weighted velocity by finding the area under the curve [which is interpreted as the displacement], then finding the constant velocity graph that has the same area. That constant velocity is the arithmetic sum of the initial and final velocities--- that is, find the rectangle with the same base that has the same area as the trapezoid.

UPDATE: Here are some v-vst-t graphs that can be used to derive the formulas for constant acceleration. (This visualizes the calculation via @BobaFit.) robphy-v-vs-t-1 robphy-v-vs-t-2

For the first graph, we find the displacement for constant-acceleration motion as the area under the v-vs-t graph, which can be interpreted as the sum of

  • the rectangle-area $v_0\Delta t$ and
  • the triangle area $\frac{1}{2} (v_f-v_0)\Delta t$,
    which can be expressed in terms of the acceleration as $\frac{1}{2} (a\Delta t)\Delta t$.

Thus, the displacement is $$\Delta x= v_0\Delta t+\frac{1}{2}a(\Delta t)^2.$$

For the second graph, the [time-weighted-] average-velocity is the "constant velocity that has the same displacement during the same time-interval", which is given by the area of the red rectangle, with height $\bar v$ and width $\Delta t$. $\bar v$ is seen to be equal to $(v_0+v_f)/2$, since \begin{align} \bar v &=\frac{\Delta x}{\Delta t}\\ &=\frac{v_0\Delta t+\frac{1}{2}(v_f-v_0)\Delta t}{\Delta t}\\ &=\frac{\displaystyle\left(\frac{v_0+v_f}{2}\right)\Delta t}{\Delta t}\\ &=\frac{v_0+v_f}{2}\\ \end{align}

  • $\begingroup$ This is fine, but you are basically using the end result in the course of the proof. I think the intention of these books is to give an integration free proof. $\endgroup$ Nov 26, 2022 at 0:24
  • $\begingroup$ @ReneSchipperus See the update with the graphical "area under the curve" derivation. You can skip the integral lines and move to the third line, using the familiar displacement formula. $\endgroup$
    – robphy
    Nov 26, 2022 at 0:25
  • $\begingroup$ Yes true, and I think this is the best approach, and is used in Tipler. But my, complaint is this intermediate equation $\overline{v}=\frac{v_0+v}{2}$ is not obvious or simpler. So I think the presentation given in these other books is faulty. $\endgroup$ Nov 26, 2022 at 0:30
  • $\begingroup$ @ReneSchipperus I agree. It seems most students don't understand these averages are time-weighted averages, not straight [equal-weighted] averages. I don't like the "displacement over elapsed time" definition.... that is, a corollary (not a definition). $\endgroup$
    – robphy
    Nov 26, 2022 at 0:33
  • $\begingroup$ You mean $\overline{v}=\frac{\Delta x}{\Delta t}$ ? This for me is the definition. But you dont take it as the definition ? $\endgroup$ Nov 26, 2022 at 0:37

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