# manipulating a physics formula

• Mar 28th 2008, 07:37 AM
vcap420
manipulating 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, 10:20 AM
wingless
$Vf^2 = Vi^2 +2a\Delta t$

Subtract $Vi^2$ from both sides

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

Divide both sides by $2\Delta t$

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

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

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,
$\Delta x$ as the change of position,
$V_1$ and $V_2$ as the first and last velocities,
$a$ as the constant acceleration

$\Delta x = \frac{V_2^2-V_1^2}{2a}$
• Mar 28th 2008, 11:34 AM
topsquark
Obviously, there was a typo in the second equation.

-Dan
• Mar 30th 2008, 07:01 AM
vcap420
thanks
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!!!