1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding instantaneous velocity at given points on nonlinear graph

  1. Jun 3, 2007 #1

    exi

    User Avatar

    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

    User Avatar

    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

    User Avatar

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

    exi

    User Avatar

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?