Deriving the kinematic equation $v^2=v_{0}^2+2a(x-x_{0})$.

Question

I have a question on deriving the kinematic equation $v^2=v_{0}^2+2a(x-x_{0})$ from first principles and the known kinematic equations. Is this simply differentiating, but with respect to time($t$)? If so, wold the process be ($a$ is a constant) $$v^2=v_{0}^2+2a(x-x_{0})\rightarrow v^2=v_{0}^2+2ax-2ax_{0}$$ $$\frac{d}{dt}\left( v^2=v_{0}^2+2ax-2ax_{0} \right)\rightarrow 2v\frac{dv}{dt}=2v_{0}\frac{dv}{dt}+2a\frac{dx}{dt}-2a\frac{dx}{dt}$$ but I am not sure if I am doing this the right way and Since I am not given a value of $\frac{dx}{dt}$ or $\frac{dv}{dt}$ I don't know which one I am looking to solve or If i am even solving the right way. I've never dealt with deriving equations with variables with subscripts, so if anyone has any insight on how to go about solving this problem, it would be greatly appreciated!

EDIT: I see that someone "complained" this question isn't about math, if this isn't math please let me know and I will delete.

Answer 1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ \begin{array}{rclcrcl} x & = & x_{0} + v_{0}t + \half\,at^{2} & \imp & x & = & x_{0} +\half\,a\pars{\color{#c00000}{t + {v_{0} \over a}}}^{2} - {v_{0}^{2} \over 2a} \\[3mm] v & = & v_{0} + at & \imp & {v \over a} & = & \color{#c00000}{t + {v_{0} \over a}} \end{array} $$

Replace $\ds{\color{#c00000}{t + {v_{0} \over a}} = {v \over a}}$ from the second equation into the first: $$ x = x_{0} + \half\,a\pars{\color{#c00000}{v \over a}}^{2} - {v_{0}^{2} \over 2a} \ \imp\ x = x_{0} + {v^{2} - v_{0}^{2} \over 2a} $$

$$ \imp\quad \color{#66f}{\Large v^{2} = v_{0}^{2} + 2a\pars{x - x_{0}}} $$

Answer 2

We don't actually need to work in terms of the kinematic formula to derive this. Note that this kinematic formula can be reorganized as $a=\dfrac{1}{2}\dfrac{\Delta (v^2)}{\Delta x}$. We can switch from this average/integral form to a differential form which can be proven directly:

$$\frac{d}{dx}\!\left(\frac{v^2}{2}\right)=v\frac{dv}{dx}=\frac{dx}{dt}\frac{dv}{dx}=\frac{dv}{dt}=a $$ where we have used the product and chain rules along with the definitions of velocity and acceleration.

Physics addendum: This can be written in terms of the kinetic energy $K=p^2/2m$ where $p=mv$ is the momentum as $$\dfrac{dK}{dx}=\frac{d}{dx}\frac{p^2}{2m}=\frac{p}{m}\dfrac{dp}{dx}=\frac{dx}{dt}\frac{dp}{dx}=\frac{dp}{dt}=F_{\text{net}}$$ where in the last equality we've used the 'momentum' statement of Newton's second law. Integrating this gives $\Delta K = \int_{x_1}^{x_2}F_{\text{net}}\,dx$ which is simply a statement of the work-energy theorem.