Equations of motion

The first two equations of motion are

v⃗av=Δd⃗Δt\displaystyle \vec{v}_{\text{av}} = \frac{\Delta{} \vec{d}}{\Delta{} t} \quad(1)anda⃗av=Δv⃗Δt\displaystyle \qquad \allowbreak\quad\allowbreak\text{and}\allowbreak\quad\allowbreak \qquad \vec{a}_{\text{av}} = \frac{\Delta{} \vec{v}}{\Delta{} t} \quad(2).

If we substitute Δx=x2−x1\displaystyle \Delta{} x = x_{2} - x_{1} for Δd⃗\displaystyle \Delta{} \vec{d} and Δv⃗\displaystyle \Delta{} \vec{v}, we get

d⃗2=d⃗1+v⃗avΔt\displaystyle \vec{d}_{2} = \vec{d}_{1} + \vec{v}_{\text{av}} \Delta{} t \quad(3)andv⃗2=v⃗1+a⃗avΔt\displaystyle \qquad \allowbreak\quad\allowbreak\text{and}\allowbreak\quad\allowbreak \qquad \vec{v}_{2} = \vec{v}_{1} + \vec{a}_{\text{av}} \Delta{} t \quad(4).

If we take Δd⃗=v⃗avΔt\displaystyle \Delta{} \vec{d} = \vec{v}_{\text{av}} \Delta{} t and substitute the average of two velocities (initial and final) for v⃗av\displaystyle \vec{v}_{\text{av}}, we get

Δd⃗=v⃗1+v⃗22Δt\displaystyle \Delta{} \vec{d} = \frac{\vec{v}_{1} + \vec{v}_{2}}{2} \Delta{} t \quad(5).

We can substitute equation (4) into this, yielding

Δd⃗=v⃗1+(v⃗1+a⃗avΔt)2Δt=2v⃗1+a⃗avΔt2Δt\displaystyle \Delta{} \vec{d} = \frac{\vec{v}_{1} + \left ( \vec{v}_{1} + \vec{a}_{\text{av}} \Delta{} t \right )}{2} \Delta{} t = \frac{2 \vec{v}_{1} + \vec{a}_{\text{av}} \Delta{} t}{2} \Delta{} t,

which simplifies to

Δd⃗=v⃗1Δt+12a⃗av(Δt)2\displaystyle \Delta{} \vec{d} = \vec{v}_{1} \Delta{} t + \frac{1}{2} \vec{a}_{\text{av}} \left ( \Delta{} t \right )^{2} \quad(6).

If we rearrange equation (4) to isolate v⃗1\displaystyle \vec{v}_{1} and then substitute that into equation (5), we get

Δd⃗=v⃗2Δt−12a⃗av(Δt)2\displaystyle \Delta{} \vec{d} = \vec{v}_{2} \Delta{} t - \frac{1}{2} \vec{a}_{\text{av}} \left ( \Delta{} t \right )^{2} \quad(7).

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 Δt\displaystyle \Delta{} t, and then we substitute that into equation (5), giving us

Δd⃗=(v⃗1+v⃗22)(v⃗2−v⃗1a⃗av)\displaystyle \Delta{} \vec{d} = \left ( \frac{\vec{v}_{1} + \vec{v}_{2}}{2} \right ) \left ( \frac{\vec{v}_{2} - \vec{v}_{1}}{\vec{a}_{\text{av}}} \right ).

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

2a⃗avΔd⃗=v⃗22−v⃗12\displaystyle 2 \vec{a}_{\text{av}} \Delta{} \vec{d} = \vec{v}_{2}^{2} - \vec{v}_{1}^{2},

which we can rearrange to get our final equation,

v⃗22=v⃗12+2a⃗avΔd⃗\displaystyle \vec{v}_{2}^{2} = \vec{v}_{1}^{2} + 2 \vec{a}_{\text{av}} \Delta{} \vec{d} \quad(8).

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.