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MisFOILing an equation?

  1. Aug 25, 2008 #1
    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 [itex]X(t)=X_0+V_0 t+\frac{1}{2}at^2[/itex]
    II [itex]V(t)=V_0+at[/itex]

    3. The attempt at a solution
    Substituting II into I gives us:
    [itex]\triangle X=V_0\frac{V_F-F_i}{a}+\frac{1}{2}a\frac{V_F-V_i}{a}^2[/itex]
    Removing the first value on the right hand side of the equals (since [itex]V_0[/itex] is typically 0), we get:
    [itex]\triangle X=\frac{1}{2}a\frac{V_F-V_i}{a}^2[/itex]
    Perhaps I've made a mistake foiling this, but I got:
    [itex]\triangle X=\frac{1}{2}a\frac{V_F^2+V_F V_i-V_F V_i+V_i^2}{a^2}[/itex]
    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:
    [itex]2a\triangle X=V_F^2+V_i^2[/itex]

    But, it's my understanding that the final answer should be:
    [itex]2a\triangle X=V_F^2-V_i^2[/itex]
    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?


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

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

    Further edit:
    Ok, since I have a [itex]V_i[/itex] 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:
    [itex]2a\triangle X=V_i*(V_F-V_i)+(V_F^2-2(V_F V_i)+V_i^2)[/itex] multiplying that first V_i through, I get:
    [itex]2a\triangle X=(V_F V_i-V_i^2)+(V_F^2-2(V_F V_i)+V_i^2)[/itex] canceling the V_i^2 I get:
    [itex]2a\triangle X=V_F V_i+V_F^2-2(V_F V_i)[/itex] then canceling out one V_F V_i I get:
    [itex]2a\triangle X=V_F^2-V_F V_i[/itex] I think.

    But [itex]V_F^2-V_F V_i[/itex] still isn't [itex]V_F^2-V_i^2[/itex] -- what am I doing wrong?
    Last edited: Aug 25, 2008
  2. jcsd
  3. Aug 26, 2008 #2


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    Homework Helper

    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 [itex]V_i*(V_F-V_i)[/itex] term)? Once you fix that I believe you'll get the answer you're looking for.

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