Equations of motion

The first two equations of motion are

${\overrightarrow{v}}_{\text{av}}=\frac{\mathrm{\Delta }\overrightarrow{d}}{\mathrm{\Delta }t}(\mathrm{𝟏})$ and ${\overrightarrow{a}}_{\text{av}}=\frac{\mathrm{\Delta }\overrightarrow{v}}{\mathrm{\Delta }t}(\mathrm{𝟐})\text{.}$

If we substitute $\mathrm{\Delta }x=x_2-x_1$ for $\mathrm{\Delta }\overrightarrow{d}$ and $\mathrm{\Delta }\overrightarrow{v}\text{,}$ we get

${\overrightarrow{d}}_2={\overrightarrow{d}}_1+{\overrightarrow{v}}_{\text{av}}\mathrm{\Delta }t(\mathrm{𝟑})$ and ${\overrightarrow{v}}_2={\overrightarrow{v}}_1+{\overrightarrow{a}}_{\text{av}}\mathrm{\Delta }t(\mathrm{𝟒})\text{.}$

If we take $\mathrm{\Delta }\overrightarrow{d}={\overrightarrow{v}}_{\text{av}}\mathrm{\Delta }t$ and substitute the average of two velocities (initial and final) for ${\overrightarrow{v}}_{\text{av}}\text{,}$ we get

$\mathrm{\Delta }\overrightarrow{d}=\frac{{\overrightarrow{v}}_1+{\overrightarrow{v}}_2}{2}\mathrm{\Delta }t(\mathrm{𝟓})\text{.}$

We can substitute equation (4) into this, yielding

$\mathrm{\Delta }\overrightarrow{d}=\frac{{\overrightarrow{v}}_1+({\overrightarrow{v}}_1+{\overrightarrow{a}}_{\text{av}}\mathrm{\Delta }t)}{2}\mathrm{\Delta }t=\frac{2{\overrightarrow{v}}_1+{\overrightarrow{a}}_{\text{av}}\mathrm{\Delta }t}{2}\mathrm{\Delta }t\text{,}$

which simplifies to

$\mathrm{\Delta }\overrightarrow{d}={\overrightarrow{v}}_1\mathrm{\Delta }t+\frac{1}{2}{\overrightarrow{a}}_{\text{av}}(\mathrm{\Delta }t{)}^2(\mathrm{𝟔})\text{.}$

If we rearrange equation (4) to isolate ${\overrightarrow{v}}_1$ and then substitute that into equation (5), we get

$\mathrm{\Delta }\overrightarrow{d}={\overrightarrow{v}}_2\mathrm{\Delta }t-\frac{1}{2}{\overrightarrow{a}}_{\text{av}}(\mathrm{\Delta }t{)}^2(\mathrm{𝟕})\text{.}$

We can derive one final equation, this time eliminating the one variable that has been present in all the others: time. We begin by rearranging equation (4) to isolate $\mathrm{\Delta }t\text{,}$ and then we substitute that into equation (5), giving us

$\mathrm{\Delta }\overrightarrow{d}=\left(\frac{{\overrightarrow{v}}_1+{\overrightarrow{v}}_2}{2}\right)\left(\frac{{\overrightarrow{v}}_2-{\overrightarrow{v}}_1}{{\overrightarrow{a}}_{\text{av}}}\right)\text{.}$

By multiplying the denominators to the other side and recognizing the difference of squares, we get

$2{\overrightarrow{a}}_{\text{av}}\mathrm{\Delta }\overrightarrow{d}={\overrightarrow{v}}_2^2-{\overrightarrow{v}}_1^2\text{,}$

which we can rearrange to get our final equation,

${\overrightarrow{v}}_2^2={\overrightarrow{v}}_1^2+2{\overrightarrow{a}}_{\text{av}}\mathrm{\Delta }\overrightarrow{d}(\mathrm{𝟖})\text{.}$

There might be a few more equations that we could have derived, but these eight (and rearranged versions of them) should take you a long way.