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

Question

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?

Answer 1

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 $$ \bar{v} = \frac{v_f+v_i}{2}\,. $$

---

For completeness, "with" calculus (although, my feeling is that the above uses calculus, too, just hidden in the geometry).

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.