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Kinematic Equations

  1. Oct 24, 2004 #1
    Since a good amount of the posts refer to problems involving kinematics I figured this may be helpful

    [tex] V_0 =[/tex] Inital Velocity
    [tex] V =[/tex] Final Velocity
    [tex] t= [/tex] time
    [tex] a= [/tex] acceleration
    [tex] x= [/tex] distance

    [tex] x=V_0t+\frac{1}{2}at^2 [/tex]

    [tex] V=V_0+at[/tex]

    [tex] V^2=V_0^2+2ax [/tex]

    [tex] x= \frac{1}{2}(V_0+V)*t [/tex]
    Last edited: Oct 24, 2004
  2. jcsd
  3. Oct 24, 2004 #2


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    Assuming that you mean "x is the distance traveled in time t" and that a is a constant, then
    [tex]x= V_0t+ \frac{1}{2}t^2[/tex]
    [tex]x= V_0t*\frac{1}{2}t^2[/tex]
  4. Oct 24, 2004 #3
    Very True... it does equal plus
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