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Kinematics homework question

  1. Sep 22, 2007 #1
    1. The problem statement, all variables and given/known data

    An object is moving in a straight line with a constant acceleration. Its position is measured at three different times, as shown in the table below.

    Time (s) Position, (m)
    37.10 7.600
    38.60 14.425
    40.10 27.100

    Calculate the magnitude of the acceleration at t=38.60 s.

    2. Relevant equations

    a = (vf - vo) / t
    d = vot + 1/2 at2
    vf2 = vo2 + 2ad
    (ether or really)

    3. The attempt at a solution

    I've tried everything, I've tried getting the distances between the 2 each time and using 1.5 seconds and getting velocity, I've tried using those to get the velocity, nothing is working. and this is something I should be able to do. Any help would be good, even just a point in the right direction.
  2. jcsd
  3. Sep 22, 2007 #2
    Welcome to the Physics Forums!

    In the equation [tex]d=v_0t+\frac{at^2}{2}[/tex]
    [tex]d[/tex] is a distance. But in the statement of the problem it says that the position of the object is measured...so you can't replace d with the data for the position.
    But if you write the distance d as the difference between the position (say [tex]x[/tex]) measured at a time and the initial position (say [tex]x_0[/tex]) and replace in the equation, then you have:


    Now all you have to do is replace the 3 sets of data in the equation above and you have a system of 3 equations with 3 unknowns (one of them being the acceleration).
    Last edited: Sep 22, 2007
  4. Sep 22, 2007 #3
    Thanks but I just got it. Before I was using 3 seconds when I was suppose to be using 1.5s, what a silly mistake.

    <I>d1</I> = 14.425-7.600 = 6.825 m / 1.5s(38.60 - 37.10) = <i>v1</I> = 4.55 m/s
    <I>d2</I> = 27.100-14.425 = 12.675 m / 1.5s = <i>v2</I> = 8.45 m/s
    <T> = 1.5s

    V1-V2/T = 2.60 m/s^2
  5. Sep 22, 2007 #4
    That is also correct.
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