It's been a while since I've been in school (10yrs). I'm a little rusty at all my math. I wonder if someone could prove how the first formula is the same as the second.

Cheers!!!

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- Mar 28th 2008, 06:37 AMvcap420manipulating a physics formula
It's been a while since I've been in school (10yrs). I'm a little rusty at all my math. I wonder if someone could prove how the first formula is the same as the second.

Cheers!!! - Mar 28th 2008, 09:20 AMwingless
$\displaystyle Vf^2 = Vi^2 +2a\Delta t $

Subtract $\displaystyle Vi^2$ from both sides

$\displaystyle Vf^2 - Vi^2 = 2a\Delta t $

Divide both sides by $\displaystyle 2\Delta t$

$\displaystyle a = \frac{Vf^2 - Vi^2}{2\Delta t}$

I see that this is not the same as the second formula you posted. Yours was $\displaystyle a = \frac{Vf^2 - Vi^2}{2\boxed{a}\Delta t}$

I don't even have an idea what formula is this. It looks like the change of position formula of a constant acceleration, without time. Which is,

$\displaystyle \Delta x$ as the change of position,

$\displaystyle V_1$ and $\displaystyle V_2$ as the first and last velocities,

$\displaystyle a$ as the constant acceleration

$\displaystyle \Delta x = \frac{V_2^2-V_1^2}{2a}$ - Mar 28th 2008, 10:34 AMtopsquark
Obviously, there was a typo in the second equation.

-Dan - Mar 30th 2008, 06:01 AMvcap420thanks
Your absolutely right there was a typo in the second equation. I was quite tired when I entered the equation. The equation involves distance rather than " t " time as well. My second lesson here is the importance of proof reading my equations. Thanks again.

Cheers!!!