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$$ \bar{v} = v_2 - \tfrac{1}{2} a \Delta t $$

$\bar{v}$ being average velocity and $v_2$ instantaneous velocity at one point

this equation was used to find instantaneous velocity from average velocity in a $\bar{v}$ vs $t$ graph, $v_2$ being the intercept in that graphic.

As $\Delta t$ tends to zero the the average velocity tends to the instantaneous velocity.

What I don't understand is where this equation came from.

I was told this equation was obtained through integrals.

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  • $\begingroup$ Hint: This formula is valid for constant acceleration. $\endgroup$ – lucas Jun 20 '16 at 4:24
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We define the average value of a quantity by : $$ \langle v \rangle=\frac{\displaystyle\int_{t_0}^{t} v ~\mathrm dt}{\displaystyle\int_{t _0}^{t} ~\mathrm dt} \ .$$

Now using the first equation of motion we get: $$v=u+at$$

Putting this in the integral we get: $$\langle v \rangle =\frac{\displaystyle\int_{t_0}^{t}(u+at)~\mathrm dt}{t-t_0}$$

which simplifies to $$\langle v \rangle =v+\frac{1}{2}a(t-t_0)\ ,$$ provided $a$ is constant.

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  • $\begingroup$ When you are using $v= u+ at ,$ it's implied that $a=\text{const.}$ $\endgroup$ – user36790 Jun 20 '16 at 5:09
  • $\begingroup$ yes, that's implied, I just provided a line at the end for clarity. $\endgroup$ – vbj Jun 20 '16 at 5:11
  • $\begingroup$ why does your equation have a positive sign and my equation a negative one? $\endgroup$ – Fernando Franco Monroy Jun 20 '16 at 8:34
  • $\begingroup$ Maybe because of your limits on $t$, or maybe your case is a deceleration. $\endgroup$ – vbj Jun 20 '16 at 9:42

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