# Simple rearranging of forumla help

1. Aug 26, 2003

### 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.

2. Aug 26, 2003

### meteor

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

3. Aug 26, 2003

### dark_exodus

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

4. Aug 26, 2003

### Integral

Staff Emeritus
s= ut + at2/2

s - ut = at2/2

2(s - ut) = at2

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

5. Aug 26, 2003

### HallsofIvy

Staff Emeritus
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.

6. Aug 26, 2003

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

7. Mar 31, 2004

### diane

rearranging equation

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

f = (1/2πC)*√(R¹+R²/R¹R²R³)

thanx

8. Mar 31, 2004

### 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:

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

?

Last edited: Mar 31, 2004
9. Mar 31, 2004

### diane

f = 1/(2πC)*√(R¹+R²
------
R¹R²R³)

plz try and help 2 rearrange 4 R² = equation

thanx

10. Apr 10, 2004

### HallsofIvy

Staff Emeritus
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.

Last edited: Apr 10, 2004
11. May 5, 2004

### r31

Worked it out thnx.

Last edited: May 5, 2004
12. Jun 14, 2004

### 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.

13. Nov 14, 2007

### 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.

14. Sep 7, 2010

### sjb-2812

Do you mean t2 at the end (as in the earlier part of this thread?)

If not, simply isolate the ut term, then divide by u. Otherwise, perhaps take s from both sides and you have a quadratic in t, which you can solve using standard methods.