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One dimensional motion

  1. Aug 30, 2004 #1


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    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)
    56.40 | 9.700
    58.20 | 19.042
    60.00 | 38.428
    Calculate the magnitude of the acceleration at t=58.20 s

    I am not quite sure how to start this. I thought you could find the velocities between the three different measurements and use this formula a =(vf-vi)/t.
  2. jcsd
  3. Aug 30, 2004 #2
    i had a question like this before. Since i didn't know calculus yet, our teacher just told us to draw the best that we could of a tangent line at t=58.2 to the curve you made with those 3 points.
    Last edited: Aug 30, 2004
  4. Aug 30, 2004 #3
    Yep, easiest way is to input that data into a graphing calculator in the table function. Then, graph that line. After that, find that point on your graph and VIOLA!

    Paden Roder
  5. Aug 30, 2004 #4


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    OK, I'll put it in my Calc and see what I get.

    Thanks PRodQuanta and needhelpperson.
  6. Aug 30, 2004 #5
    oops, i forgot that a constant acceleration has a linear graph with velocity vs time. you can just figure out the change in velocity over the first 2 points and divide it over the change in time. sorry bout that...
  7. Aug 30, 2004 #6


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    Thanks needhelpperson that worked and really helped. I will remember that for next time.
  8. Aug 31, 2004 #7


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    I think if you realize that [tex]x = x_0 + v_0 t - \frac{a t^2}{2}[/tex] then you can use your data to obtain a simple linear system of equations for the three unknowns [tex]x_0[/tex], [tex]v_0[/tex] and [tex]a[/tex]. Then you can use your solution of this system to calculate the acceleration or whatever you want at any time!
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