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Simple rearranging of forumla help

  1. Aug 26, 2003 #1

    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:
  2. jcsd
  3. Aug 26, 2003 #2
    The correct formula is
  4. Aug 26, 2003 #3
    Could you give a break down of how you got that answer?
  5. Aug 26, 2003 #4


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    s= ut + at2/2

    s - ut = at2/2

    2(s - ut) = at2

    a = 2(s - ut))/(t2)
  6. Aug 26, 2003 #5


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    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.

    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.
  7. Aug 26, 2003 #6
    please excuse my dear aunt sally , she dosn't know her order of operations.
  8. Mar 31, 2004 #7
    rearranging equation

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

    f = (1/2πC)*√(R¹+R²/R¹R²R³)

  9. Mar 31, 2004 #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]


    [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]

    Last edited: Mar 31, 2004
  10. Mar 31, 2004 #9
    f = 1/(2πC)*√(R¹+R²

    plz try and help 2 rearrange 4 R² = equation

  11. Apr 10, 2004 #10


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    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.
    Last edited by a moderator: Apr 10, 2004
  12. May 5, 2004 #11


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    Worked it out thnx.
    Last edited: May 5, 2004
  13. Jun 14, 2004 #12


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    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.
  14. Nov 14, 2007 #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.
  15. Sep 7, 2010 #14
    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.
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