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Derive the following kinematics equation using the position and velocity equations

  1. Dec 25, 2012 #1
    1. The problem statement, all variables and given/known data

    [tex]{ y }_{ f }={ y }_{ i }+{ v }_{ yi }t+\frac { 1 }{ 2 } { a }_{ y }{ t }^{ 2 }\\ { v }_{ yf }={ v }_{ yi }+{ a }_{ y }t[/tex]

    Derive

    [tex]{ { v }_{ yf } }^{ 2 }={ { v }_{ yi } }^{ 2 }+2{ a }_{ y }({ y }_{ f }-{ y }_{ i })[/tex]

    2. The attempt at a solution

    This isn't an actual homework question; I'm just cross-posting this thread because I'm terribly impatient and there is much more traffic here.
     
  2. jcsd
  3. Dec 25, 2012 #2
    Re: Derive the following kinematics equation using the position and velocity equation

    Solve for t in the second equation, and substitute it into the first equation.
     
  4. Dec 25, 2012 #3
    Re: Derive the following kinematics equation using the position and velocity equation

    Thanks. That derivation took much longer than expected.
     
    Last edited: Dec 25, 2012
  5. Dec 25, 2012 #4

    ehild

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    Homework Helper
    Gold Member

    Re: Derive the following kinematics equation using the position and velocity equation

    You can connect the equation to work-energy theorem. The work done on a particle is equal to the change of its kinetic energy. Work done is displacement times force. Force is F=ma. [tex]W=(y_f-y_i)ma=1/2(mv_f^2-mv_i^2)[/tex]. Cancel m.

    But you can derive the formula easily if you remember that the displacement is average velocity multiplied by time. [tex]Δy=\frac{1}{2}(v_i+v_f)t[/tex]. Substitute [itex]t=\frac{v_f-v_i}{a}[/itex] for t.
    [tex]y_f-y_i=\frac{1}{2}(v_i+v_f)\frac{v_f-v_i}{a}=\frac{v_f^2-v_i^2}{2a}[/tex]

    ehild
     
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