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Kinematic Equation vs. Delta velocity / Delta Time - Factor of 2?

  1. Dec 6, 2012 #1
    I am testing a cylindrical piece of steel dropping down steel pipe in a vacuum system, and came across a problem comparing the kinematic equation and a standard change in velocity over change in time to determine an acceleration. My derivation shows that the kinematic equation is exactly 2 times that of delta v/ delta time.

    I have attached my calculations. Has anyone came accross this issue before?

    It is driving me crazy! Thanks for the help.
     

    Attached Files:

  2. jcsd
  3. Dec 6, 2012 #2

    K^2

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    Science Advisor

    Ah, because (d2-d1)/(t2-t1) gives you average velocity between t1 and t2. Not the instantaneous velocity at t2, which is what you need for your formula to work. If object is at rest at t1, under uniform acceleration, velocity at t2 will be exactly twice the average velocity between t1 and t2. That might be the factor of 2 you are looking for.
     
  4. Dec 6, 2012 #3

    Doc Al

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    Staff: Mentor

    Realize that (d2-d1)/(t2-t1) is the average velocity between t1 and t2. It's not the the speed at time t2.

    The average velocity (for uniform acceleration) = (Vi + Vf)/2.
     
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