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Linearizing quadratic equation

  1. Sep 19, 2007 #1
    1. The problem statement, all variables and given/known data

    basically im suppose to linearize d=vit + 1/2at^2 into d/t^2 vs 1/t, t/d vs t^2/d and 1/d vs 1/t
    and its asking which graphs would be linear and non-linear.

    2. Relevant equations

    d=vit + 1/2at^2

    3. The attempt at a solution

    all i know is that in order to linearize this i need to make it into the format y=mx+b from there i can see which variables alternate and which remain constant.
    now im confused on how im suppose to get d/t^2 vs 1/t for the first one so i started to solve for t but i ended up needing to use the quadratic equation since its ax^2 + bx + c=0 but that really didnt help me since i coudlnt isolate "d" again. So basically im kinda lost on how to manipulate this formula.
  2. jcsd
  3. Sep 19, 2007 #2


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    For the first part, I'd let y = d/t^2, let x = 1/t. Can you get the d and t equation into an equation with y and x... substitute appropriately and manipulate the formulas.
  4. Sep 19, 2007 #3
    If you divide the equation by t^2 you get

    [tex]\frac{d}{t^2} = \frac{v_i}{t} + \frac{1}{2}a[/tex]

    If you wanted that against 1/t, basically make x a substitution of 1/t, you would have a graph that looked pretty linear with a slope of vi.

    Do you see how it works?
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