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.5at^2 Why .5?

  1. Jan 22, 2010 #1
    1. The problem statement, all variables and given/known data
    Knowing and understanding the kinematic equation's variables x1, x0, v0, t, and a, how has .5 been incorporated into the equation?

    I realized this unknown during lab. Now, as I've nearly completed my analysis, I find that the values I've derived for acceleration should be doubled!


    2. Relevant equations
    x1 = x0 + (v0 * t) + (.5 * a * t^2)


    3. The attempt at a solution
    By using dimensional analysis one can determine that each side of the equation yields distance.
     
  2. jcsd
  3. Jan 22, 2010 #2

    rock.freak667

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    Constant acceleration

    [tex]v= \int a dt = at+C[/tex]

    t=0,v=v0


    [tex]x=\int v dt = \int (v_0 +at) dt = v_o t + \frac{1}{2}at^2 +C_1[/tex]

    t=0,x=x0⇒C1=x0

    x=x0+v0t+½at2

    how exactly do your values show that the '½' should be '2'?
     
  4. Jan 22, 2010 #3
    I'm using Δx/Δt/Δt to produce a value for acceleration. When that value is applied to the kinematic equation, it returns value that does not coincide with the experimental result. By removing the .5, the answer is correct.
     
  5. Jan 22, 2010 #4

    ideasrule

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    That's because Δx/Δt/Δt is not acceleration. Δx/Δt is not the maximum speed attained in Δt; it's the average speed, which is half the maximum speed (assuming constant acceleration). Acceleration would be 2Δx/Δt/Δt.
     
  6. Jan 22, 2010 #5

    Redbelly98

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    That is incorrect. The acceleration is Δv/Δt.

    However, Δv is not the same as Δx/Δt, because: Δv is the change in velocity, while Δx/Δt is the average velocity.

    For an easy example, consider the case of a constant (but nonzero) velocity. Clearly Δv and a are both zero. However Δx/Δt is not zero since the object is moving at constant velocity.
     
  7. Jan 22, 2010 #6

    Redbelly98

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    If one is starting from rest, yes that is true.
     
  8. Jan 23, 2010 #7
    I was incorrectly using a vavg for Δv. I've changed my velocity calculations to reflect an instantaneous velocity, and so my calculations for acceleration are correct and my kinematic equation produces a correct result without any modification.
     
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