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Homework Help: Tangent Lines/Instaneous Velocity

  1. Sep 23, 2013 #1
    1. The problem statement, all variables and given/known data
    I need to use the following data table to:
    1. make a position-time graph
    2. draw tangents (it isn't specified how many, but a previous practice question only drew 3 out of a possible 5)
    3. create a time-velocity table.

    0 --- 0
    0.25 --- 0.29
    0.50 --- 1.15
    0.75 --- 2.59
    1.00 --- 4.60
    1.25 --- 7.19
    1.50 --- 10.35

    2. Relevant equations
    vinst = d/t

    3. The attempt at a solution
    The position-time graph is attached. I think that's right, but what I'm really having trouble with is drawing the tangent lines and calculating the instantaneous velocity. I feel like I didn't draw the tangents right but I'm not sure how to do it differently.
    Here's my calculations:
    0.50: 7 / 1.75-0.25 = 3.7
    1.00: 10 / 1.65-0.50 = 5.5
    1.50: 10 / 1.50-0.75 = 6.2

    Attached Files:

  2. jcsd
  3. Sep 23, 2013 #2

    Andrew Mason

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    A tangent line at a certain point on the graph is a straight line that touches the graph only at that point.

    What does the slope of the tangent line on a distance-time graph represent? Does that help you determine points for a velocity-time graph?

  4. Sep 23, 2013 #3
    It seems to me you did an excellent job, and have the right idea. Nice work.
  5. Sep 24, 2013 #4

    Andrew Mason

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    Your tangent lines (supposedly tangent at .5, 1.0 and 1.5) are not touching at those points. This affects the calculation of velocity at those points. If you plot the velocity vs. time you will see this. I would choose a tangent at t=1.25 since you do not have any points after t=1.5.

  6. Sep 24, 2013 #5
    Thanks for the help.
    Do you mean do a tangent at 1.25 instead of at 1.50 or do both?
  7. Sep 24, 2013 #6
    Gee. It looks to me like they are pretty tangent at these points. The tangent lines aren't perfect, but they look pretty good. Yet, when I calculated the velocities using central finite difference approximations, I got a straight line as a function of time, with all the velocities calculated from the drawn tangents lying below the (more accurate) numerically calculated velocities.

  8. Sep 24, 2013 #7

    Andrew Mason

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    The problem with drawing a tangent at 1.5 is that you don't know what the line looks like after. I would use 1.25 but you can use as many as you like: you can use any points in between 0 and 1.5.

  9. Sep 24, 2013 #8

    Andrew Mason

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    Actually, the real problem is with your calculations. The slope of the tangent you have drawn at t=1.0 is 10/(1.65-.50) = 9.1 not 5.5. The slope of the tangent as you have drawn it at 1.5 is 10.35/(1.5-.75) = 13.8 (the point is (10.35, 1.5)). Check the slope calculation for t=.5. That should give you an accurate v-t graph.

    You seem to be dividing by the first term in the denominator and then subtracting the second!! Slope is rise/run

    Last edited: Sep 24, 2013
  10. Sep 24, 2013 #9
    Ah I see, thank you!
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