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?

  • $\begingroup$ 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. $\endgroup$
    – Void
    Commented Sep 21, 2014 at 9:47
  • $\begingroup$ Related: physics.stackexchange.com/q/55809/2451 $\endgroup$
    – Qmechanic
    Commented Sep 21, 2014 at 19:29

2 Answers 2


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 $v_{avg}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.