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Graph Problem

  1. Oct 9, 2006 #1
    I am having trouble figureing out the y = mx + b of this if the table of values look like

    Distance Time (s)
    25.000 0.000
    24.920 1.000
    24.678 2.000
    24.269 3.000
    23.685 4.000
    22.913 5.000
    21.932 6.000
    20.712 7.000
    19.209 8.000
    17.349 9.000
    15.000 10.000
    11.874 11.000
    7.000 12.000
    0.000 13.000

    I am tring to find out the y = mx + b of this curve and then I have to figure out the derivative of the curve? Any help please?
     
  2. jcsd
  3. Oct 9, 2006 #2

    berkeman

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    Are those (x,y) values? If so, what do you have for y when x=0? If you put that point into y=mx+b, what does that point tell you about b?

    If you graph the points, do you get a straight line?
     
  4. Oct 9, 2006 #3

    quasar987

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    Well if you look at the position as a function of time and write

    x(t)=mt+b,

    you should at least be able to find what b is by looking at the first set of data:

    Distance Time (s)
    25.000 0.000

    At t=0.000, x=25.000.

    A little effort.
     
  5. Oct 9, 2006 #4
    I am having trouble getting the slope? Because the graph is a curve and the y value decreases as time increases.
     
  6. Oct 9, 2006 #5

    Integral

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    I will be interested in seeing what line you choose to draw through that data.
     
  7. Oct 9, 2006 #6

    berkeman

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    Yeah, no kidding. I just graphed it in Excel, and she's not a straight line. Not even close.

    So to the original poster (OP), what *exactly* is the problem statement?
     
  8. Oct 9, 2006 #7

    quasar987

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    Is this the data for the free fall of an object?

    2000 posts! *champagne and trumpets*
     
  9. Oct 9, 2006 #8
    This is the question A 25 ft long ladder is resting against a vertical wall and the bottom of the ladder is moving out at a fixed rate of 2.00 ft per second. Graph x and y as functions of time. and I did x which is easy and the y = mx + b was y = 1/2x + 0 I just need some help with y?
     
  10. Oct 9, 2006 #9

    radou

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    Well, x(t) is of the form mx +b, precisely, x(t) = v*t + x0, where the velocity of the bottom of the ladder is given, as should the initial position x0 be. Now simply use the Pythagora's theorem x^2+y^2 = 25^2, and express y out of it. After doing this, plug your function x(t) into the obtained equation, to get the function y(t).
     
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