# How to prove the average velocity formula without calculus

I often see the formula:

$$\overline v = \frac {v + v_0}{2}$$

in physics textbooks, to describe the average velocity for an object moving with constant acceleration. This formula is often substituted into the $x = x_0 + \overline v t$ formula in order to derive the $x = x_0 + v_0t + 1/2at^2$ formula we are so familiar with.

With a calculus background we don't even need this formula. We just keep integrating from a = constant until we get the same formula. But if you don't have a calculus background I guess using this first equation is a necessary stepping stone on the way to deriving the formula.

That the average velocity is the mid-point between the initial and final velocities makes intuitive sense I guess, but I want to be able to demonstrate to students that this is true in a more rigorous way. The only problem is that the only way I can see to proving that this formula is indeed true is to use calculus!

In particular, I would use the knowledge that the area under the v/t curve is the displacement. Drawing a simple v/t graph that is linear (constant acceleration) I could show geometrically that the area under this curve is the same formula.

And there's the problem. The area under the curve bit of knowledge implies a knowledge of calculus.

Is there any other way to demonstrate to a non-calculus student that it is true? I've never seen any textbook attempt to justify this formula. It just seems to pop out of nowhere without any rigorous proof.

My take...

First, consider constant velocity $v$, then by daily intuition, the average velocity $\bar v=v$ (because its the same $v$ throughout, thus the average cannot be anything other than $v$)

Now consider acceleration, and the initial velocity $v_0$, Then the final velocity $v$ must be greater than $v_0$

The average $\bar v$, which is some kind of representative of the initial and final velocities, thus must obey the following:

$$v_0<\bar v<v$$

as if $v_0=\bar v$ or $v = \bar v$, then it does not represent that there's a contribution of $v$ (respectively $v_0$) in it

Since we have two velocities $v$ and $v_0$, we expect the representative $\bar v$ must be contributed equally by them. This suggest that $v$ and $v_0$ are in equal proportions to made up $\bar v$

$$\bar v= nv+nv_0$$

Now it is easy to see that $0<n<1$ as otherwise the above inequality will be violated

Therefore having $n=\frac{1}{2}$ is a solution

(Thanks wikipedia for the better answer below!)

2) Being a representative also means it must be a value that balances the given values thus the following must also be true (That is, the total difference between the representative to each of the given velocities must vanishes else you will end up having $\bar v$ being tipped over towards either $\bar v$ or $v$ more):

$$(v-\bar v)+(v_0-\bar v)=0$$

Thus rearranging gives:

$$\frac{v_0+v}{2}=\bar v$$

As required

3) A geometric proof • Thanks for the answer, I like the clear explanation. It's obvious to me that if we treat the velocities as a set of data points that the average is v + v_0 /2. I still think that more background is required in order to link this to the total displacement, ie splitting the area under the curve into smaller and smaller rectangles, concept of limits, etc. – esotechnica Nov 30 '15 at 1:22

$v(t)$ is a line. This should be an OK assumption. So it's of the form $mt+b$. If you consider the interval $t_1$ through $t_2$ and draw a horizontal line of height $\frac{1}{2}(v(t_1)+v(t_2))=m \frac{1}{2}(t_1+t_2)+b$, you'll see that exactly half of the velocity line lies above the horizontal line, and exactly half lies below. Therefore that must be the average.
If the "exactly half above, exactly half below" argument doesn't hold (for discrete numbers that's the median, not the mode), discretize the line by $N$ points and prove the average of that is always $\frac{1}{2}(v(t_1)+v(t_2))$. I.e., prove
$$\frac{1}{N+1}\sum_{n=0}^N v(t_n)= \frac{1}{2}(v_0+v_f)$$