MisFOILing an equation?

  • Thread starter Banaticus
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In summary, the conversation is discussing how to derive a third common equation using two given equations. The attempt at a solution involves substituting one equation into the other and simplifying, but there is a mistake in foiling which leads to an incorrect answer. The edit points out the mistake and corrects it, but there is still a missing factor in the equation. The correct equation should be 2aΔX=V_i*(V_F-V_i)+1/2(V_F^2-2(V_F V_i)+V_i^2). Once this is corrected, the final answer is 2aΔX=V_F^2-V_i^2.
  • #1
Banaticus
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Edit: See edit at the bottom of the post

Homework Statement


From a couple of common equations, derive a third common equation.

Homework Equations


I [itex]X(t)=X_0+V_0 t+\frac{1}{2}at^2[/itex]
II [itex]V(t)=V_0+at[/itex]

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?

Edit:

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:
[itex]V_F^2-V_i^2[/itex]

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:
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  • #2
Hi Banaticus,

Banaticus said:
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]

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.

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?
 
  • #3


I would suggest carefully checking your algebra steps to ensure that you are accurately simplifying the equation. It appears that you may have made a mistake in your foiling and simplification process. It may also be helpful to double check the original equations and their variables to ensure that you are substituting them correctly. If you are still having trouble, I would recommend seeking assistance from a classmate or your instructor. Remember to always check your work and ask for help when needed.
 

1. What does it mean to "MisFOIL" an equation?

The term "MisFOIL" refers to an error made while using the FOIL method to multiply two binomials. It usually involves mixing up the order of the terms or incorrectly distributing the negative sign.

2. How can I identify if an equation has been MisFOILed?

If an equation has been MisFOILed, you will typically see terms that do not match the correct FOIL pattern. For example, instead of (a + b)(c + d), you may see something like (a + c)(b + d), which is incorrect.

3. What are the consequences of MisFOILing an equation?

MisFOILing an equation can lead to incorrect solutions and misinterpretation of results. It can also make the equation more difficult to solve and can result in a waste of time and resources.

4. How can I avoid MisFOILing an equation?

To avoid MisFOILing an equation, it is important to carefully follow the correct FOIL method and double-check your work. It can also be helpful to simplify the equation before attempting to FOIL.

5. Can MisFOILing an equation be fixed?

Yes, MisFOILing an equation can be fixed by identifying the error and correcting it. This may involve rearranging terms or distributing negative signs correctly. It is important to carefully check your work after making corrections to ensure the equation is now correctly FOILed.

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