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Homework Help: Finding instantaneous velocity at given points on nonlinear graph

  1. Jun 3, 2007 #1

    exi

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    1. The problem statement, all variables and given/known data
    Problem is presented as a nonlinear graph. Y-axis is in meters and ranges from 0.0 to 40.0m in increments of 10m, and the X-axis is in seconds, 0.0-5.0s (increments of 1). Points on graph include (0,10), (1,~19), (2, ~23), (3, ~26), (4, ~26), and (5, ~20).

    Question: "Estimate the instantaneous velocity of the train at 2.0s." (and again at 4.0s).

    2. Relevant equations
    V(av) = delta(x) / delta(t)
    V(as delta(t) approaches 0) = delta(x) / delta(t).

    3. The attempt at a solution
    Unsure.

    I understand the concept behind finding instantaneous velocity in that it's an instant measure of motion and includes a directional component, but damn if I can remember how to do it. The notes from class are a bit too complicated to be of use, and guesstimating the slope of a line tangent to x=2 isn't much help, given this graph.

    Any thoughts?
     
  2. jcsd
  3. Jun 3, 2007 #2
    Do you have a copy of the graph online? You should be able to obtain a reasonable tangent by using a ruler and a pencil, and drawing your tangent onto your graph.
     
  4. Jun 3, 2007 #3

    exi

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    I'm afraid I don't. Besides drawing in more precise increments on the axes, sketching the tangent line, and taking the slope of that where it intersects the point in question, how should I be going about this? All of the questions before that one on these notes are in reference to distance, displacement, average speed and velocity - all the usual stuff - but I'm still a little unsure as to the most efficient way to solve it.
     
  5. Jun 3, 2007 #4
    well at 2s it's zero
     
  6. Jun 3, 2007 #5

    exi

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    And how is that?
     
  7. Jun 3, 2007 #6
    im sorry i meant 4s
     
  8. Jun 3, 2007 #7

    exi

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    Alright... but how is that?
     
  9. Jun 3, 2007 #8
    (3, ~26), (4, ~26) because dx/dt is zero there? as in it doesn't move as in it's position doesn't change, again if i could see the graph it would be much easier to tell.

    i guess it could have a steep negative slope around there somewhere.
     
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