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Is this a valid Kinematics Equation?

  1. Dec 12, 2015 #1
    a = dv/dt
    a = (d/dt)(dx/dt)
    dv/dt = (d/dt)(dx/dt)
    dv = dx/dt
    dx = dvdt
    Xf - Xi = (tf-t0)*dv

    Is this a valid equation? It doesn't seem right to me since if velocity change is 0, then dv = 0, meaning your change in position is 0. That shouldn't be right, as even though your velocity change is 0, you still have velocity meaning you are moving. But the math seems like it works out.
     
  2. jcsd
  3. Dec 12, 2015 #2

    Orodruin

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    This step is wrong. Integrating both sides gives ##v = dx/dt## which is the definition of ##v##.
     
  4. Dec 12, 2015 #3
    Oh, right! I should have read dv/dt as (d/dt)(v).
     
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