# Motion under constant acceleration, is the proof correct?

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?

• 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. Nov 26, 2022 at 0:04

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.)

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}

• 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. Nov 26, 2022 at 0:24
• @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. Nov 26, 2022 at 0:25
• 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. Nov 26, 2022 at 0:30
• @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). Nov 26, 2022 at 0:33
• You mean $\overline{v}=\frac{\Delta x}{\Delta t}$ ? This for me is the definition. But you dont take it as the definition ? Nov 26, 2022 at 0:37