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Rearranging equations.

  1. Jul 20, 2005 #1
    In 19 days I will be taking a physics class. I have already taken two algebra classes yet I do not know how to rearrange equations. What are the rules for rearranging an equation. F=m*a, how do you rearrange it so you know what m= ?
     
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  3. Jul 20, 2005 #2

    HallsofIvy

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    You "undo" whatever is done to a. In F= ma, a is multiplied by m. To "undo" that, you do the opposite of "multiply by m"- you divide by m. Dividing both sides of the equation by m, F/m= (ma)/m= a so a= F/m.
     
  4. Jul 20, 2005 #3

    quasar987

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    mmmh.. by 'algebra' I guess you mean 'arithmetics' ?


    The key idea is that F, m and a represent numbers, like 1, 2, 3¼, pi³,...etc. And for any number 'n', there exists another number 'm' such that n*m = 1. This number m is of course, the inverse of n: m = 1/n.

    So if for exemple, we want to isolate m in F=ma, we want to multiply both sides by the inverse of a, like so

    [tex]F=ma \Leftrightarrow \frac{1}{a}F=m\frac{a}{a} \Leftrightarrow \frac{F}{a} = m[/tex]
     
  5. Jul 20, 2005 #4
    t=(Vf-Vi)/a if I wanted to find "a" I would multiply both sides by a, yes? but that would leave me with at=Vf-Vi right? and "a" would not be isolated on one side of the =. what am I doing wrong?
     
  6. Jul 20, 2005 #5

    quasar987

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    Continue. Divide both sides by t.
     
  7. Jul 20, 2005 #6

    TD

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    In these type of problems you only need to know two fundamental properties of equations. Equation remain equivalent under two certain operations.


    a) An eqaution remains equivalent when you add the same number to both sides (this may be negative if you wish to 'substract').

    Example, we want a out of: [tex]a + b = c \Leftrightarrow a + b - b = c - b \Leftrightarrow a = c - b[/tex]

    b) An eqaution remains equivalent when you multiply both sides with the same factor ([tex] \ne 0[/tex])

    Example, we want a out of: [itex]a\frac{b}{c} = d \Leftrightarrow a\frac{b}{c}\frac{c}{b} = d\frac{c}{b} \Leftrightarrow a = \frac{{dc}}{b}
    [/itex]

    That's all you need here :smile:
     
  8. Jul 20, 2005 #7
    thanks everyone :biggrin:
     
  9. Jul 20, 2005 #8

    EnumaElish

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    An eqaution remains equivalent when you multiply both sides with the same factor, including 0.

    Generally, an eqaution remains equivalent when you apply an invertible (strictly increasing or strictly decreasing) function to both sides:

    a = b implies Log(a) = Log(b);

    a = sqrt(b) implies a2 = (sqrt(b))2 = b.

    In general, if a = f(b) and g is the inverse of f, then g(a) = g(f(b)) = b.
     
    Last edited: Jul 20, 2005
  10. Jul 20, 2005 #9

    TD

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    I thought that in equivalent equations, the solutions have to remain the same.
    In that case, you have to exclude 0, no?
     
  11. Jul 20, 2005 #10
    ok I have the equation a=(V2-V1)/t , And I want to make it V2=? . it would be a*V2= (-V1)/t ,is it good so far? but i dont know what to do with "a"
     
  12. Jul 20, 2005 #11

    EnumaElish

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    You have to pay attention to the parantheses.

    a = (V2 - V1)/t
    at = V2 - V1
    at + V1 = V2
     
  13. Jul 20, 2005 #12

    EnumaElish

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    I see what you're saying. I am splitting hairs when I point out that strictly speaking, a = b is preserved under multiplication with zero, which is not covered under your (or, the)definition of equivalent equations.
     
    Last edited: Jul 20, 2005
  14. Jul 20, 2005 #13
    why did you out everything except V2 on the other side of the = instead of putting V2 were "a" was and put "a" on the other side?

    why did you multiply "t" and "a"?
    if "t" was the numerator, would you still multiply it with "a"?
     
  15. Jul 21, 2005 #14

    ek

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    I remember in grade 11 my chem teacher taught us some little triangle method for the equation n=cv to solve for each quantity.

    The the next year in Phys12 I sit down at my desk the first day and written on my desk is the little triangle cheat method and someone commented below it "morons use triangles". I always got a kick out of that.

    How old are you and what grade are you in? I didn't take a physics class until grade 10 and we were doing stuff like this in grade 7/8. I find it odd that you would be taking a physics class without knowing these algebra fundamentals.
     
  16. Jul 21, 2005 #15

    ek

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    why did you out everything except V2 on the other side of the = instead of putting V2 were "a" was and put "a" on the other side?

    It's the same thing. V = at and at = V are the same thing.

    why did you multiply "t" and "a"?

    You multiply both sides by t. The t's cancel on the right side and you're left with at = V2 - V1

    if "t" was the numerator, would you still multiply it with "a"?

    No, then you would divide it out. Divide both sides by t and you're left with a/t = V2-V1 (Which obviously isn't a valid equation)
     
  17. Jul 21, 2005 #16

    EnumaElish

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    ek's explanation is right:

    a = (V2 - V1)/t
    (V2 - V1)/t = a
    t(V2 - V1)/t = ta
    V2 - V1 = ta
    V2 - V1 + V1 = ta + V1
    V2 = V1 + ta
     
  18. Aug 13, 2005 #17
    I've got a problem I don't know how to rearrange. d=Vt+1/2 at^2 ,i'm supposed to find out what t equals.
    d/V=Vt/V+1/2 at^2
    d/V=t +1/2 at^2
    2d/V=t +2*1/2 at^2
    2 d/V=t + at^2
    (2 d/V)/a=t + at^2/a
    (suare root of)(2 d/V)/a=t + (square root of)t^2
    (suare root of)(2 d/V)/a=t +t
    ((suare root of)(2 d/V)/a)/2=2t/2
    ((suare root of)(2 d/V)/a)/2=t
    What did I do wrong? how do I fix it so Im not dividing fractions
     
  19. Aug 13, 2005 #18
    What do you mean "at power of two"? Do you mean the whole expression is to the power of 2?

    And is Vt one variable or two?
     
  20. Aug 13, 2005 #19

    VietDao29

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    You are wrong at the second line.
    You have:
    [tex]a = b + c[/tex]
    [tex]\Leftrightarrow \frac{a}{d} = \frac{b + c}{d} = \frac{b}{d} + \frac{c}{d}[/tex]
    We divide both sides by d <> 0.
    Anyway, to find t from:
    [tex]d = vt + \frac{1}{2}a t ^ 2[/tex]
    [tex]\Leftrightarrow \frac{a}{2}t ^ 2 + vt - d = 0[/tex]
    t is the unknown.
    Can you solve:
    [tex]\alpha x ^ 2 + \beta x + \gamma = 0[/tex]
    for x?
    [tex]\alpha , \beta , \gamma[/tex] are already known.
    Viet Dao,
     
  21. Aug 13, 2005 #20
    (ax^2)/a+bx-bx+y-y=(-bx-y)/a

    x^2=(-bx-y)/a

    square root both sides

    x= square root of (-bx-y)/a


    That's really messy and I don't know if thats how you would do it.
     
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