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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?

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  • $\begingroup$ 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 Sep 21 '14 at 9:47
  • $\begingroup$ Related: physics.stackexchange.com/q/55809/2451 $\endgroup$ – Qmechanic Sep 21 '14 at 19:29
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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.

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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 $v_{avg}$.

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