# How do you arrive at this equation? $\bar{v} = v_2 - \tfrac{1}{2} a \Delta t$

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

$\bar{v}$ 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.

• Hint: This formula is valid for constant acceleration. – lucas Jun 20 '16 at 4:24

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.
• When you are using $v= u+ at ,$ it's implied that $a=\text{const.}$ – user36790 Jun 20 '16 at 5:09
• Maybe because of your limits on $t$, or maybe your case is a deceleration. – vbj Jun 20 '16 at 9:42