Simple rearranging of forumla help

  • Thread starter dark_exodus
  • Start date
In summary, the formula for rearranging s=ut+1/2at2 to find t is by isolating the ut term and then dividing by u, or by taking s from both sides to create a quadratic in t that can be solved using standard methods.
  • #1
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. :frown:
 
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  • #2
The correct formula is
s=(v*t)+((1/2)*a*t*t)
Then
a=(s-(v*t))/((1/2)*t*t)
 
  • #3
Could you give a break down of how you got that answer?
 
  • #4
s= ut + at2/2

s - ut = at2/2

2(s - ut) = at2


a = 2(s - ut))/(t2)
 
  • #5
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
please excuse my dear aunt sally , she dosn't know her order of operations.
 
  • #7
rearranging equation

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

f = (1/2πC)*√(R¹+R²/R¹R²R³)

thanx
 
  • #8
You're going to have to put in more parantheses in that expression in the radical... Do you mean:

[tex]R^1 + \frac{R^2}{R^1R^2R^3}[/tex]

or:

[tex]R^1 + \frac{R^2}{R^1}R^2R^3[/tex]

or perhaps:

[tex]\frac{R^1 + R^2}{R^1R^2R^3}[/tex]

?
 
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  • #9
f = 1/(2πC)*√(R¹+R²
------
R¹R²R³)

please try and help 2 rearrange 4 R² = equation

thanx
 
  • #10
That's even worse!

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

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
[tex]\frac{2f}{\pi C}= \sqrt{\frac{R_1+R_2}{R_1R_2R_3}}[/tex]

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

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

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

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

You are welcome to go back to superscripts now if that was the way the problem was given.
 
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  • #11
Worked it out thnx.
 
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  • #12
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
Is it possible someone could help me rearrange this equation to find [tex]g[/tex] please?

[tex] T = 2\pi \sqrt (\frac{\ell}g)[/tex]

Heres what I've done.

[tex] T = 2\pi \sqrt (\frac{\ell}g)[/tex]

Divide both sides by [tex]2\pi[/tex]

[tex]\frac{T}{2\pi} = \sqrt (\frac{\ell}g)[/tex]

Remove the root by squaring both sides

[tex]\frac{T^2}{4\pi^2} = \frac{\ell}g[/tex]

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

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

I've got the figures for [tex]T[/tex], [tex]\ell[/tex] and [tex]g[/tex]. If I use that final rearrangement I can get closer to the answer for [tex]g[/tex] 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
JoeCFC1310 said:
Hi I need help rearranging s=ut+1/2at2 to find t
any help would be appreciated

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.
 

What is simple rearranging of formula?

Simple rearranging of formula is the process of changing the order of variables or terms in a mathematical equation to solve for a specific variable or simplify the equation.

Why is simple rearranging of formula important?

Simple rearranging of formula is important because it allows scientists to solve complex equations and determine relationships between different variables. It also helps in making predictions and analyzing data.

What are the steps for simple rearranging of formula?

The steps for simple rearranging of formula include identifying the variable to be solved for, isolating that variable on one side of the equation, and rearranging the equation to solve for the variable.

Are there any rules or guidelines for simple rearranging of formula?

Yes, there are certain rules and guidelines to follow when rearranging formulas, such as using inverse operations, keeping equations balanced, and being careful with negative signs and exponents.

Can simple rearranging of formula be used in all types of equations?

Yes, simple rearranging of formula can be used in most types of equations, including linear equations, quadratic equations, and even more complex equations involving multiple variables. However, it may not always be possible to rearrange an equation to solve for a specific variable.

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