Simple rearranging of forumla help

[dark_exodus]

Hi

I need to rearrange the following formula to get a(Acceleration):

s = ut-1/2at

The last t is squared

Any help
My maths is a little rusty. [:(]

[meteor]

The correct formula is
s=(v*t)+((1/2)*a*t*t)
Then
a=(s-(v*t))/((1/2)*t*t)

[dark_exodus]

Could you give a break down of how you got that answer?

[Integral]

s= ut + at2/2

s - ut = at2/2

2(s - ut) = at2


a = 2(s - ut))/(t2)

[HallsofIvy]

It's pretty basic algebra:

You have the formula s = ut-1/2at2 and want to solve for a. To "solve for a" means to change it into an equation like
a= something.

You do that using two basic concepts: (1) For everything that is alread "done" to a, do the opposite (2) Anything you do to one side of the equation you must do to the other.

So:
s= ut- (1/2) at2. a is not "by itself" because it has been multiplied by -(1/2)t2 and has ut added to itself. The opposite of adding ut is subtracting ut: subtracting ut from both sides gives
s- ut= ut- (1/2) at2- ut = -(1/2) at2

Now the only problem is that a is multiplied by -(1/2) t2.
So, divide both sides of the equation by -(1/2) t2.

That gives (s- ut)/((-1/2)t2)= a so, after a little simplifying, that the value of a:

a= -2(s-ut)/t2.

[On Radioactive Waves]

please excuse my dear aunt sally , she dosn't know her order of operations.

[diane]

hiya, can u help me rearrange the following equation 2 get R² -

f = (1/2πC)*√(R¹+R²/R¹R²R³)

thanx

[Muzza]

You're going to have to put in more parantheses in that expression in the radical... Do you mean:

R^1 + \frac{R^2}{R^1R^2R^3}

or:

R^1 + \frac{R^2}{R^1}R^2R^3

or perhaps:

\frac{R^1 + R^2}{R^1R^2R^3}

?

[diane]

f = 1/(2πC)*√(R¹+R²
------
R¹R²R³)

plz try and help 2 rearrange 4 R² = equation

thanx

[HallsofIvy]

That's even worse!

I'm going to assume that "f = (1/2πC)*√(R¹+R²/R¹R²R³)" means
f= (\frac{1}{2}\pi C)\sqrt{\frac{R_1+R_2}{R_1R_2R_3}}

Notice that I have also changed to sub-scripts rather than super-scripts since I tend to confuse those with exponents (I am assuming they are NOT exponents!).

First thing you do is divide both sides by that number outside the square root to get
\frac{2f}{\pi C}= \sqrt{\frac{R_1+R_2}{R_1R_2R_3}}

Now get rid of that square root by squaring both sides:
\frac{4f^2}{\pi^2C^2}= \frac{R_1+R_2}{R_1R_2R_3}

Multiply on both sides by R1R2R3 so we don't hav e that fraction to worry about:
(R_1R_2R_3)\frac{4f^2}{\pi^2C^2}= R_1+ R_2

Subtract R2 from both sides so that we have the quantity we are solving for on the left:
(R_1R_2R_3)\frac{4f^2}{\pi^2C^2}- R_2= R_1
and, since there is an "R2" in each term, factor that out:
R_2((R_1R_3)\frac{4f^2}{\pi^2C^2}- 1)= R_1

Finally, isolate R2 by dividing both sides of the equation by everything on the left except R2:
R_2= \frac{R_1(\pi^2C^2-1)}{R_1R_3\pi^2C^2}

You are welcome to go back to superscripts now if that was the way the problem was given.

[r31]

Worked it out thnx.

[H20]

What the hell? Mulliday, I think you should stop posting rubbish. I've seen several stupid posts by you! Please take time to sit down and think deeply and carefully about the following question: Why are you spamming this forum? Think very carefully about it, and try and make everything that comes to your mind into a focused point. Then, see whether you really think your points are sensible. That should help you.

[Matty R]

Is it possible someone could help me rearrange this equation to find g please?

T = 2\pi \sqrt (\frac{\ell}g)

Heres what I've done.

T = 2\pi \sqrt (\frac{\ell}g)

Divide both sides by 2\pi

\frac{T}{2\pi} = \sqrt (\frac{\ell}g)

Remove the root by squaring both sides

\frac{T^2}{4\pi^2} = \frac{\ell}g

This is where I'm a bit confused. Does the rearrangement finish as :

\ell divided by (\frac{T^2}{4\pi^2}) = g ?

I've got the figures for T, \ell and g. If I use that final rearrangement I can get closer to the answer for g than any other, but my answer is slightly larger than the given answer.

I would appreciate any help at all, and I apologise for any mistakes in my coding. Its the first time I've ever used Tex.