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Finding the second equation

Now it's time to find the second equation. This gets a little trickier than the first because substitution will be needed to solve, so I hope your algebra skills are up to par.

First off, we have to assume a few more equations that pretty much just commen sense. These two equations are:

Vaverage=(Vf+Vi)/2
Vaverage=(xf-xi)/(tf-ti)

Again, these two equations both make sense given that you know what a velocity is.

Now, if any of you have ever taken basic algebra, then you know what I'm about to do. Since both of the equation equal Vaverage, then we can set these two equations equal to each other. This will give us:

(xf-xi)/t=(Vf+Vi)/2

Now we want to solve for xf, so use your algebra skills and follow along.

xf-xi=t((Vf+Vi)/2)
xf=t((Vf+Vi)/2)+xi

Now what I'm about to do may not make sense, but I'll try to explain it the best I can. I'm going to take out the Vf and insert the first equation in its place. This rids the equation of a variable and allows you to solve for distance much easier. So here's how it looks:

xf=t((at+Vi+Vi)/2)+xf
xf=((at2+2tVi)/2)+xi
xf=1/2at2+Vit+xi

This is our final equation. You can now solve for final distance by only knowing the initial position, the acceleration, the initial velocity, and the total time. Only one equation left to go. If you're brave enough, you can try looking at the third equation.