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How to prove v(t) = (y(t+Δt) - y(t-Δt))/2Δt

  1. Feb 8, 2014 #1
    1. Prove that:
    v(t) = (y(t+Δt) - y(t-Δt))/2Δt

    using kinematic equations for constant acceleration.


    3. I tried using the limit as Δt approaches 0 of (y(t+Δt)-y(t))/Δt and somehow adding to get Δt. Could I manipulate the kinematic equations somehow by setting t equal to (t+Δt) or (t-Δt)?
     
  2. jcsd
  3. Feb 8, 2014 #2

    PeroK

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    What's the normal definition of v(t) in terms of these sorts of limits?
     
  4. Feb 8, 2014 #3

    Andrew Mason

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    Since acceleration is constant: y(t+Δt) = y(t) + vavgΔt = y(t) + (v(t) + v(t+Δt))Δt/2

    Does that help?

    AM
     
  5. Feb 8, 2014 #4
    Try using the equation:

    y(t)=y(0)+v(0)t+at2/2

    Evaluate y(t+Δt) and y(t-Δt), subtract them, and divide by 2Δt, and see what you get.
     
  6. Feb 8, 2014 #5

    PeroK

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    Perhaps:

    y(t)=y(0)+v(t)t-at2/2

    might help as well.
     
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