Deriving the 4th equation of motion

[danksnaks]

1. All I need to do is derive the 4th equation of motion being v²= v0²+2aD from the second (t=(v-v0)/a) and third (D=1/2at²+v0t).
2. In this case D= (Ending-initial distance) V0= Initial Veolocity
3. By having the second equation solved for t I could substitute it in an completely eliminate time and I end up with D=1/2a((v-v0)/a))²+v0((v-v0)/a). Past this point I knwo to distribute the terms, but I have no idea what the correct forms look like because I always ended up with a slight different in the term hooked to 1/2a. Any help here?

 

[danksnaks]

I lied. I figured it out.

 

[tomcue]

I understand your question and I can provide some guidance on how to derive the 4th equation of motion.

First, let's start with the second equation of motion, t=(v-v0)/a. We can rearrange this to solve for v, giving us v=v0+at.

Next, we can substitute this expression for v into the third equation of motion, D=1/2at^2+v0t. This gives us D=1/2a(v0+at)^2+v0(v0+at).

Expanding the terms and simplifying, we get D=1/2a(v0^2+2v0at+a^2t^2)+v0^2+av0t.

Now, we can combine like terms and factor out a v0, giving us D=1/2a(v0^2+2av0t+a^2t^2)+v0(v0+at).

Finally, we can rearrange this equation to solve for v^2, giving us v^2=v0^2+2aD. This is the 4th equation of motion that you were looking for.

I hope this explanation helps and clarifies the steps needed to derive the 4th equation of motion. Remember, it's important to practice and work through the equations to fully understand them. Keep up the good work!