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Finding Acceleration Experimentally

  1. Oct 8, 2009 #1
    You drop an object. You take the time it takes to hit the ground. So you have Δx (displacement) and Δx (time).

    To find acceleration do you use

    xf= xi + Vi t + .5at2
    Δ x= Vi t + .5at2
    Δ x- Vi t=.5at2
    a= 2(Δ x- Vit )/ t2

    or do you use

    v= Δ x/ Δ t
    a= Δ v/ Δ t

    Should acceleration come out the same? Why do they differ so much? Take example a Δx of 2m and a time of .5. Using the first equation you get 16 m/s^2. But using the second method, you get 8 m/s^2.
  2. jcsd
  3. Oct 8, 2009 #2


    Staff: Mentor

    Use the first method. The second method is wrong. The key problem is that v=Δx/Δt is an average velocity, not a change in velocity. So you cannot turn around and plug v into the spot for Δv in the second expression. In other words, an average v is not a Δv. In fact, in this case the change in velocity is twice the average velocity (that should not be too surprising) which is where the factor of 2 comes from.
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