TITLE: Getting the kinematic equation $v^2=v_0^2+2a(x−x_0)$
QUESTION [0 upvotes]: As I've searched through the internet, looking for explanation. I found an answer with transformation of this formula:
$$
x = x_0 + v_0t + \frac{1}{2}at^2 ⟹ x = x_0 + \frac{1}{2}a(t + \frac{v_0}{a})^2 - \frac{v^2_0}{2a} 
$$
I don't understand how to get $\frac{1}{2}a(t + \frac{v_0}{a})^2$
Can anyone help please?

REPLY [0 votes]: Here is a formulation which gives a role to the average velocity. Let $v_{avg}:=\dfrac{\Delta x}{\Delta t}=\dfrac{x-x_0}{t}$ be the average velocity from time $0$ to $t$. Since this is constant-acceleration motion, we may use the kinematic equation for position to write
$$v_{avg} = \frac{v_0 t+\frac12 at^2}{t}=v_0+\frac{at}{2}.$$
By contrast, the kinematic equation for velocity gives the final velocity as $v=v_0+at$. As such, we may write $v_{avg}=(v+v_0)/2$---that is, the average velocity at constant acceleration is also the average of the initial and final velocities. Invoking constant acceleration once more, we have
$$a=\frac{\Delta v}{\Delta t}=\frac{\Delta v}{\Delta x}\frac{\Delta x}{\Delta t} = \frac{\Delta v}{\Delta x}v_{avg}=\frac{v-v_0}{x-x_0}\frac{v+v_0}{2}=\frac{v^2-v_0^2}{2(x-x_0)}$$
which is the desired equation. (Note how closely this parallels the use of the chain rule, which would apply even if $a$ weren't constant.)
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