Kinematic Equations:
\begin{align} v &= v_0+at \\ \Delta x &= \left( \frac {v+v_0} 2 \right) t \\ \Delta x &= v_0t + \frac12 at^2 \\ v^2 &= v_0^2 + 2a\Delta x \end{align}
So for the fourth equation, final velocity squared is found without $t$, time.
My question: is it possible for this equation to be derived or explained visually/graphically?
For the first three equations it is easy to come up with a visual representation as to why the equations make sense under constant acceleration. I tried to imagine the 4th equation as a $v^2$-vs.-$t$ graph, that is quadratic. But my mind can’t understand how $$2a\Delta x$$ can be explained.
I understand how the equation is derived algebraically, but my question is whether the equation can be easily explained in geometric terms like the other kinematic formulas.
One way that you could visualise this is by using the graph $a$ vs $\Delta x$. Since $a\Delta x$ has the same units as $v^2$ this means that the area under an acceleration vs displacement graph from $0$ to $\Delta x$ has to be equal to the area under the graph velocity vs velocity from $v_0$ to $v$.
In kinematics the acceleration is constant so the area under the $a$ vs $\Delta x$ graph is simply $a\Delta x$ meanwhile the area under the velocity graph would be the area of the trapezium of side lengths $v_0$ and $v$ and height $v-v_0$.
$$a\Delta x=(v-v_0)({v+v_0\over2})$$ $$a\Delta x={v^2-v_0^2\over2}$$ $$v^2={v_0}^2+2a\Delta x$$ or more properly: $$\int_0^{\Delta x}adx=\int_{v_0}^vvdv$$ $$a(\Delta x-0)={v^2\over2}-{{v_0}^2\over2}$$ $$v^2={v_0}^2+2a\Delta x$$ Hope this is what you were looking for.