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Can any1 explain vf^2 = vi^2 + 2ad?

  1. Oct 30, 2005 #1

    can anyone explain the logistics of vf^2 = vi^2 + 2ad to me? i mean its one thing to simply remember the equation but, to understand how they came about it is way more helpful.

    thx alot :biggrin:
  2. jcsd
  3. Oct 30, 2005 #2


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    Gold Member

    Here is the proof:
    1) [tex] X = \frac{V_{f} + V_{i}} {2} \Delta t
    you can prove this with an integral. (constant acceleration)
    2) [tex] a = \frac{V_{f} - V_{i}}{ \Delta t }
    3)[tex] \Delta t = \frac{V_{f} - V_{i}}{ a }
    4) [tex] X = \frac{V_{f} + V_{i}} {2} \cdot \frac{V_{f} - V_{i}}{ a }

    5) [tex] X = \frac{V_{f}^2 - V_{i}^2} {2 a}

    6) [tex] V_{f}^2 = V_{i}^2 + 2 a X
    Last edited: Oct 30, 2005
  4. Oct 30, 2005 #3


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    Staff: Mentor

    Here's another way. Start with the general equations for position and velocity under constant acceleration:
    [tex]x = x_0 + v_0 t + \frac{1}{2} a t^2[/tex]
    [tex]v = v_0 + a t[/tex]
    Make the following substitutions: [itex]x_0 = 0[/itex], [itex]x = d[/itex], [itex]v_0 = v_i[/itex] and [itex]v = v_f[/itex].
    Solve the two equations together to eliminate [itex]t[/itex], then rearrange to solve for [itex]v_f^2[/itex].
  5. Oct 30, 2005 #4
    Thx alot

    ohhhh now i c,
    why didnt i see that?
    thanks alot :biggrin:
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