# MisFOILing an equation?

1. Aug 25, 2008

### Banaticus

Edit: See edit at the bottom of the post

1. The problem statement, all variables and given/known data
From a couple of common equations, derive a third common equation.

2. Relevant equations
I $X(t)=X_0+V_0 t+\frac{1}{2}at^2$
II $V(t)=V_0+at$

3. The attempt at a solution
Substituting II into I gives us:
$\triangle X=V_0\frac{V_F-F_i}{a}+\frac{1}{2}a\frac{V_F-V_i}{a}^2$
Removing the first value on the right hand side of the equals (since $V_0$ is typically 0), we get:
$\triangle X=\frac{1}{2}a\frac{V_F-V_i}{a}^2$
Perhaps I've made a mistake foiling this, but I got:
$\triangle X=\frac{1}{2}a\frac{V_F^2+V_F V_i-V_F V_i+V_i^2}{a^2}$
The inner and outer terms drop out and a positive squared and a negative squared are both a positive, right? So, then after canceling an "a" and moving the remaining "a" and the 1/2 over to the left hand side:
$2a\triangle X=V_F^2+V_i^2$

But, it's my understanding that the final answer should be:
$2a\triangle X=V_F^2-V_i^2$
Note the - between the velocity values, not a +, which I got. So, what am I doing wrong? Is the equation supposed to be a - not the + which I think it should be? Did I make a mistake foiling?

Edit:

Ok, I did make a mistake foiling. The middle terms shouldn't drop out. They aren't:
$V_F^2+V_F V_i-V_F V_i+V_i^2$ but rather:
$V_F^2-V_F V_i-V_F V_i+V_i^2$ which simplifies to:
$V_F^2-2(V_F V_i)+V_i^2$

Still, how do I get this to become: $V_F^2-V_i^2$? I could drop out anything with $V_i$ in it, since we assume that itt's zero and anything multiplied by zero is zero, but then I'd up with just a:
$V_F^2$ instead of a:
$V_F^2-V_i^2$

Further edit:
Ok, since I have a $V_i$ in the final answer, I probably shouldn't have dropped it to begin with. If I pull the 1/2, cancel an a and move the rest over to the other side of the problem (and remembering that V_0 and V_i are the same thing), I think I end up with:
$2a\triangle X=V_i*(V_F-V_i)+(V_F^2-2(V_F V_i)+V_i^2)$ multiplying that first V_i through, I get:
$2a\triangle X=(V_F V_i-V_i^2)+(V_F^2-2(V_F V_i)+V_i^2)$ canceling the V_i^2 I get:
$2a\triangle X=V_F V_i+V_F^2-2(V_F V_i)$ then canceling out one V_F V_i I get:
$2a\triangle X=V_F^2-V_F V_i$ I think.

But $V_F^2-V_F V_i$ still isn't $V_F^2-V_i^2$ -- what am I doing wrong?

Last edited: Aug 25, 2008
2. Aug 26, 2008

### alphysicist

Hi Banaticus,

This equation is missing one factor. Originally, the second term on the right had a factor of 1/2, which you got rid of by multiplying both sides by 2. So after that, now the term on the left is multiplied by 2 (so that is correct), but what about the first term on the right (the $V_i*(V_F-V_i)$ term)? Once you fix that I believe you'll get the answer you're looking for.