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Acceleration (of an orange, 2 possible answers)

  1. Feb 18, 2012 #1
    1. The problem statement, all variables and given/known data
    We are given information about a falling orange, in a data table regarding its position, in a position time graph, from the graph of the position time graph, we have found instantaneous velocity of the orange at certain time intervals, but finding the slopes of tangents at those time intervals. Using these velocities, we graph a velocity time graph, and are asked to find the acceleration of the orange as it falls to the ground.



    2. Relevant equations


    a=v2-v1/t2-t1


    3. The attempt at a solution


    for v1 and t1 should i use 0? since right at the beginning v1 and t1 = 0

    or should i use the very first registered values of the orange falling ( first time interval = 0.5 s and the velocity of the orange at that time = 5.33 m/s )

    I come up with 2 possible answers for acceleration.

    a= v2-v1/t2-t1

    9.14-0/1-0

    =9.14/1

    a=9.14 m/s

    OR

    =9.14-5.33/1-0.5

    =3.81/0.5

    a=7.62 m/s


    which of these 2 answers make sense?

    Thanks!
     
  2. jcsd
  3. Feb 19, 2012 #2

    PhanthomJay

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    You can use either method, but you are getting different answers because you apparently either calculated the the instantaneous velocities incorrectly from your distance vs. time graph, or the data contains experimental errors. The acceleration should be close to g, assuming air resistance is small over this distance.
     
  4. Feb 19, 2012 #3
    If the body starts from rest take ##v_1=0##, if the body is in motion and stops take ##v_2=0##, ##v_1## or ##v_2##, is the velocity of the body you would have
    ##a=\frac{Δ v}{Δ t}## if you have constant velocity ##a=\frac{v}{t}##
     
    Last edited: Feb 19, 2012
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