# Derivation of the average Velocity formula with constant acceleration (using calculus)

I have been looking at

$$v_{avg} = \frac{v_{i} + v_{f}}{2},$$

when the acceleration is constant, where $v_i$ is equal to the initial velocity and $v_f$ is equal to the final velocity.

How can you derive this using calculus?

$$v_{avg} = \frac{\Delta x}{\Delta t} = \frac{1}{t_{f} - t_{i}} \int_{t_{i}}^{t_{f}} v(t) dt$$

We know that

$$v(t) = at + v_0,$$

thus by substituting this into the previous integral yields

$$v_{avg} = \frac{1}{t_{f} - t_{i}} \left[a\frac{t_f^2}{2} + v_{0} t_f - a\frac{t_i^2}{2} - v_{0} t_i\right].$$

But there's nothing else, which I can think of. Idea?

• I have edited the maths to make it clearer, if you meant something else by your notation (or if you don't like the edit), you can roll back the changes. – Void Sep 21 '14 at 9:47
• – Qmechanic Sep 21 '14 at 19:29

There is one error in the derivation, if you want to have $v(t_i)=v_0$, you must have $$v(t) = v_0 + a(t-t_i)$$ You also have to use the fact that $v_f = v(t_f)$. Once you use all this, you should be able to divide out $t_f-t_i$ in the corrected version of your last line and get the result you seek.
You are almost done. If you simplify the express you found for $v_{avg}$ and compare it with the initial equation for $v_{avg}$, in which you have to substitute $v_i$ for $v(t_i)$ and $v_f$ for $v(t_f)$, you will see that they are equivalent.
Another way of deriving this is by expressing $v(t)$ in terms of $v_i$ and $v_f$:
$$v(t) = v_i + \frac{v_f - v_i}{t_f - t_i} (t - t_i)$$
Simplifying the resulting expression of the integral of this equation should also give you your initial equation for $v_{avg}$.