If we take
Δd=vavΔt
and substitute the average of two velocities (initial and final) for
vav,
we get
Δd=2v1+v2Δt(5).
We can substitute equation (4) into this, yielding
Δd=2v1+(v1+aavΔt)Δt=22v1+aavΔtΔt,
which simplifies to
Δd=v1Δt+21aav(Δt)2(6).
If we rearrange equation (4) to isolate
v1
and then substitute that into equation (5), we get
Δd=v2Δt−21aav(Δt)2(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, and then we substitute that into equation (5), giving us
Δd=(2v1+v2)(aavv2−v1).
By multiplying the denominators to the other side and recognizing the difference of squares, we get
2aavΔd=v22−v12,
which we can rearrange to get our final equation,
v22=v12+2aavΔd(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.