# Proof 14: Motion equations

In motion along a path, $a$ is acceleration, $v_0$ and $v$ are initial and final (current) speed, and $t_0$ and $t$ are initial and final (current) time.

#### The 1st motion equation starts from $a=v’$:

$$\begin{align}

~&a=v’=\frac{ \mathrm dv}{ \mathrm dt}\\

\Leftrightarrow\quad & a \,\mathrm dt= \mathrm d v\\

\Leftrightarrow\quad & \int a \,\mathrm dt= \int \mathrm d v\qquad\leftarrow \text{ Assumed start at time } t_0=0 \text{, since integral has no limits}\\

\Leftrightarrow\quad & at+c_1=v+c_2\qquad\leftarrow a \text{ assumed constant}\\

\Leftrightarrow\quad & v=at+\underbrace{c}_{c_1-c_2}\\

\Leftrightarrow\quad & v=v_0+at\quad_\blacksquare\qquad\leftarrow c \text{ is initial speed } v_0 \text{ at } t=0 \text{, since: } v=a\cdot 0+c=c

\end{align}$$

#### The 2nd motion equation starts from $v=s’$ and inputs the 1st equation:

$$\begin{align}

~&v=s’=\frac{ \mathrm ds}{ \mathrm dt}\\

\Leftrightarrow\quad & v \,\mathrm dt= \mathrm d s\\

\Leftrightarrow\quad & \int \underbrace{v_0+at}_v \,\mathrm dt= \int \mathrm d s\qquad\leftarrow \text{ All assumptions on } v=v_0+at \text{ are now added}\\

\Leftrightarrow\quad & v_0t+\frac12 at^2+c_1=s+c_2\\

\Leftrightarrow\quad & s=v_0t+\frac12 a t^2+\underbrace{c}_{c_1-c_2}\\

\Leftrightarrow\quad & s=s_0+v_0t+\frac12at^2\quad_\blacksquare\qquad\leftarrow c \text{ is start location } s_0 \text{ at } t=0

\end{align}$$

#### The 3rd motion equation starts from the 2nd equation and inputs the 1st:

$$\begin{align}

s&=s_0+v_0t+\frac12 at^2\\

~&=s_0+v_0\underbrace{\left(\frac{v-v_0}a\right)}_t+\frac12 a \underbrace{ \left(\frac{v-v_0}a\right)^2 }_{t^2} \qquad\leftarrow\text{ Insert }v=v_0+at \text{, assuming } a\neq 0\\

~&=s_0+\frac{vv_0-v_0^2}a+\frac12 \cancel a \frac{(v-v_0)^2}{a^{\cancel 2}}\\

~&=s_0+\frac{vv_0-v_0^2}a+\frac12 \frac{v^2+v_0^2-2vv_0}a\qquad\leftarrow \text{ Square of binomial (see } \href{https://www.allthatmatters.dk/proof-15-square-of-binomial-and-a-sum-times-a-difference/}{\text{Proof 15}} \text{)}\\

~&=s_0+\frac{vv_0}a-\frac{v_0^2}a+\frac12 \frac{v^2}a+\frac12 \frac{v_0^2}a-\frac1 {\cancel 2} \frac{\cancel 2vv_0}a\\

~&=s_0+\cancel{ \frac{vv_0}a}-\cancel{ \frac{vv_0}a }-\frac{v_0^2}a+\frac12 \frac{v_0^2}a +\frac12 \frac{v^2}a \\

~&=s_0-\frac12 \frac{v_0^2}a+\frac12 \frac{v^2}a\\

\Leftrightarrow & \quad s-s_0=\frac{v^2}{2a}- \frac{v_0^2}{2a} \\

\Leftrightarrow & \quad (s-s_0)2a=v^2- v_0^2 \\

\Leftrightarrow & \quad v^2 = v_0^2+ 2a (s-s_0)\quad_\blacksquare \\

\end{align}$$

#### The 4^{th} motion equation also starts from the 2^{nd} equation and inputs the 1^{st}:

$$\begin{align}

s&=s_0+v_0t+\frac12 at^2\\

~&=s_0+v_0t+\frac12 \left(\frac{v-v_0}{\cancel t}\right)t^{\cancel 2}\qquad\leftarrow \text{ Insert } v=v_0+at\\

~&=s_0+v_0t+\frac12 (v-v_0)t\\

~&=s_0+v_0t+\frac12 vt-\frac12 v_0t\\

~&=s_0+\frac12 v_0t+\frac12 vt\\

s&=s_0+\frac12 (v+v_0)t \quad_\blacksquare

\end{align}$$

All four motion equations require constant acceleration and $t_0=0$.

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