Getting the kinematic equation $v^2=v_0^2+2a(x−x_0)$

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.)

Considering that $a$ is constant, we can approach this problem alternatively as follows begin{align*} a(s_{1} - s_{0}) = int_{s_{0}}^{s_{1}}amathrm{d}s = int_{s_{0}}^{s_{1}}frac{mathrm{d}v}{mathrm{d}t}mathrm{d}s = int_{s_{0}}^{s_{1}}frac{mathrm{d}s}{mathrm{d}t}mathrm{d}v = int_{s_{0}}^{s_{1}}vmathrm{d}v = frac{v^{2}(s_{1}) - v^{2}(s_{0})}{2} end{align*}

and the result holds.

hint

$$At^2+Bt=A(t^2+frac{Bt}{A})$$

with

$$t^2+frac{Bt}{A}=(t+frac{B}{2A})^2-(frac{B}{2A})^2$$

Replace $ A $ by $ frac 12a $ and

$B $ by $ v_0$.