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How was this formula derived?

  1. Sep 17, 2011 #1
    1. The problem statement, all variables and given/known data
    I'm trying to derive this formula, but I get stuck after I factor the t out.

    V[itex]_{f}[/itex][itex]^{2}[/itex] = V[itex]_{0}[/itex][itex]^{2}[/itex] + 2a (y-y[itex]_{0}[/itex])

    2. Relevant equations

    V[itex]_{f}[/itex][itex]^{2}[/itex] = V[itex]_{0}[/itex][itex]^{2}[/itex] + 2a (y-y[itex]_{0}[/itex])

    3. The attempt at a solution
    1) y[itex]_{f}[/itex] - y[itex]_{0}[/itex] = ([itex]\frac{V_{0}+V_{f}}{2}[/itex])t

    2) y[itex]_{f}[/itex] - y[itex]_{0}[/itex] ([itex]\frac{1}{t}[/itex])= ([itex]\frac{V_{0}+V_{f}}{2}[/itex])t ([itex]\frac{1}{t}[/itex])

    3) V[itex]_{f}[/itex]= ([itex]\frac{V_{0}+V_{f}}{2}[/itex])t ([itex]\frac{1}{t}[/itex])

    4) ????

    5) V[itex]_{y}[/itex] = V[itex]_{0y}[/itex][itex]^{2}[/itex] + 2a (y[itex]_{f}[/itex]-y[itex]_{0}[/itex])
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 17, 2011 #2

    PeterO

    User Avatar
    Homework Helper

    Firstly, I can only show this using symbols
    v for your Vf - final velocity
    u for your Vo - initial velocity
    s for our y - yo - displacement.
    {partially from familiarity, and partly because it is easier to type}

    So I am aiming at v2 = u2 + 2as

    you are hopefully familiar with a couple of other motion equations

    v = u +at & s = t*(v + u)/2 [you listed tis second one in line 1 of your solution]

    These two are combined.

    The first can be transposed to give

    t = (v - u)/a

    substitute for t in the second

    s = (v-u)(v+u)/2a

    so

    (v-u)(v+u) = 2as

    v2 - u2 = 2as

    or

    v2 = u2 + 2as
     
  4. Sep 17, 2011 #3
    Alternatively:

    a = dv/dt

    a(ds) = ds/dt(dv)

    int(ads) = int(vdv)

    Assuming a = constant.

    as = 1/2(v^2-u^2)

    v^2 = u^2 + 2as
     
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