# With constant acceleration, can $\bar{v}=\Delta{x}/\Delta{t}=(v_i+v_f)/2$ be established rigorously without calculus?

I have found no way to rigorously establish the basic one-dimensional kinematic equation relating the two following expressions of average velocity without using some kind of argument that amounts to taking limits as is done in calculus

$$\bar{v}=\frac{\Delta{x}}{\Delta{t}}=\frac{v_i+v_f}2 .$$

I can provide the following heuristic argument.

By definition constant velocity means $$\Delta x=v\Delta t$$. We can illustrate that as either a point-slope graph, or as the area under the line representing constant velocity.

From that we can illustrate $$\bar{v}\Delta{t}=\Delta{x}$$ in the same way, by saying average velocity multiplied by the time interval gives the displacement $$\Delta x$$ as if it were a constant velocity.

The actual position and velocity graphs for constant acceleration are more complicated. At this point in the development I can't even justify the graph of position with respect to time. But I can hand-wave it. Even without the $$x:t$$ graph it is intuitively clear that the position at $$t=t_o+\Delta t/2$$ isn't $$x_o+\Delta x/2.$$ That's because the body moves further in the second half than in the first half of the period.

By definition the average of the two numbers $$v_o$$ and $$v_f$$ is half their sum. If, based on the case of constant velocity, we assume the area under the velocity graph is the displacement, then simple geometry and algebra give

\begin{align} \Delta x&=x_f-x_o\\ &=\left(v_o+\frac{v_f-v_o}{2}\right)\Delta t\\ &=\frac{v_o+v_f}{2}\Delta t\\ &=\bar{v}\Delta t\\ &=\left(v_o+\frac{a\Delta t}{2}\right)\Delta t,\\ x_f&=x_o+v_o\Delta t+\frac{1}{2}a\Delta t^2. \end{align}

Without using the methods of calculus, I see no way to obtain these results without either assuming

$$\bar{v}=\frac{v_o+v_f}{2}=\frac{\Delta x}{\Delta t},$$

or assuming the area under the velocity graph equals the displacement. Is it possible to rigorously show the two forms $$\bar{v}=\Delta{x}/\Delta{t}$$ and $$\bar{v}=\left(v_i+v_f\right)/2$$ of average velocity are equal without using calculus?

• You are using instantaneous velocities, aren't you? How do you define these $v_0$ and v $v_f$ used in your formulas withut calculus? It seems that you wave your hand even before starting the actual hand-waving.
– nasu
Commented Oct 16, 2023 at 23:15
• "assuming the area under the velocity graph equals the displacement" really isnt an assumption; with a few smartly chosen examples, before-even-pre-calculus students can easily be persuaded that this is true. And then you can use the argument you had. The thing that really is difficult to motivate, is "why constant acceleration?", and that the concept of instantaneous velocities, and acceleration itself, are calculus-derived. We can, however, study "piecewise affine-linear functions", first in s-t v.s. v-t graphs, then try them in v-t to see what happens in s-t, thereby getting the quadratics. Commented Oct 17, 2023 at 4:41
• @naturallyInconsistent I agree that it's a fairly easy sale to persuade someone that area under the velocity vs time graph is displacement. But the only way to make the claim rigorous seems to be proving the FTC. IMO instantaneous velocity is an oxymoron. But I also believe the concept of "infinitesimal" is indispensable. Commented Oct 18, 2023 at 18:29
• For piecewise horizontal velocities, there is neither a problem with instantaneous velocities nor does anything here require calculus. Showing that it should be area under velocity time graph in those situations does not require FTC. The special case of piecewise linear velocities giving rise to quadratic dependences upon time had proofs that came before modern rigorous calculus, i.e. before FTC. You can proceed with either just considering the consequences of piecewise linear velocities, of which the areas are trivial to find, or you can bring in just derivatives. Commented Oct 19, 2023 at 1:25

If we define the average velocity as the $$\bar{v} = \frac{\Delta x}{\Delta t} = \frac{x_f-x_i}{t_f-t_i}\,,$$ then we can use the constant-acceleration kinematic equation $$x_f-x_i = v_i(t_f-t_i) +\frac{1}{2}a(t_f-t_i)^2$$ to compute this as $$\bar{v} =\frac{x_f-x_i}{t_f-t_i} =\frac{v_i(t_f-t_i) +\frac{1}{2}a(t_f-t_i)^2}{t_f-t_i} =v_i +\frac{1}{2}a(t_f-t_i)\,.$$ From another kinematic equation, $$v_f = v_i + a (t_f-t_i)\,,$$ this can be simplified to $$\bar{v} = \frac{v_f+v_i}{2}\,.$$
We define the average velocity as the time-weighted velocity. That is, \begin{align} \bar{v} &= \frac{1}{t_f-t_i}\int_{t_i}^{t_f}v(t)dt\,. \end{align} In general, we can compute this as \begin{align} \bar{v} &= \frac{1}{t_f-t_i}\int_{t_i}^{t_f}v(t)dt \\&= \frac{1}{t_f-t_i}\int_{t_i}^{t_f}x'(t)dt \\&= \frac{1}{t_f-t_i}(x(t_f)-x(t_i)) \\&= \frac{\Delta x}{\Delta t}\,, \end{align} which means that it has the same meaning as the constant velocity one would travel to get $$\Delta x$$ displacement in $$\Delta t$$ time.
For constant acceleration, the time-weighted average velocity can be computed as \begin{align} \bar{v} &= \frac{1}{t_f-t_i}\int_{t_i}^{t_f}v(t)dt \\&= \frac{1}{t_f-t_i}\int_{t_i}^{t_f}(v_i+at)dt \\&= \frac{1}{t_f-t_i}\left(v_i(t_f-t_i)+\frac{1}{2}a(t_f-t_i)^2\right) \\&= v_i+\frac{1}{2}a(t_f-t_i) \\&= v_i+\frac{1}{2}(v_f-v_i) \\&= \frac{v_f+v_i}{2}\,. \end{align} Thus, under the definition of the average velocity $$\bar{v}$$ as the time-weighted velocity, we have generally that $$\bar{v} = \Delta x/\Delta t$$ and that for the special case of constant-acceleration motion, $$\bar{v}$$ is equal to the average of the initial and final velocities.
• @StevenThomasHatton (I missed the word "without" in the very title of your post, so, mea culpa). But I'm not sure what the issue is. If you define the average velocity as the time-weighted velocity, then you have an integral, by definition. If, on the other hand, you start with the definition that the average velocity is $\Delta x/\Delta t$, i.e., $\bar{v}$ is the constant velocity that gets you $\Delta x$ in $\Delta t$, then $(v_f+v_i)/2$ follows directly from the kinematic equations: $\Delta x/\Delta t = (v_i\Delta t+\frac{1}{2}a\Delta t^2)/\Delta t = (v_f+v_i)/2$ via steps in my post. Commented Oct 16, 2023 at 22:21
• @StevenThomasHatton Reading your post more closely, I think the answer to your question is no. You're right that you need to assume that the area under the velocity curve is the total displacement to derive the kinematic equation for $x_f-x_i$, and I think that this "is" calculus. Commented Oct 16, 2023 at 22:28