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Position - time graph

  1. Aug 2, 2013 #1
    Hey guys, so I have a quick question about position - time graphs, so without using calculus for lets just say a particle moving west along a horizontal straight line every 0.10 s and the displacement is 0.022, 0.032, 0.042, etc.

    Well the displacement from t = 0 aren't really good numbers that I used lol, but lets just assume it's non - linear, and without calculus would we just use a tangent line and find the slope of the tangent line to figure out velocity at what ever time interval we are given.


    Thanks
     
  2. jcsd
  3. Aug 2, 2013 #2
    I think that is correct, the slope at that particular point is your velocity.
     
  4. Aug 2, 2013 #3
    That's what I'm thinking but I wasn't certain.
     
  5. Aug 4, 2013 #4

    collinsmark

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    Be careful. There are two different measures of velocity, and either one might be asked for, depending on the problem.

    In both cases, assume that you already have a displacement vs. time curve.

    • The tangent of any point on the curve gives you the instantaneous velocity. That's the velocity of the particle at a particular instant in time.
    • But if you're given a particular time interval (meaning two, separate points on the curve), it usually means you are being asked to find the average velocity. For that, use

      [tex] \vec {v_{\mathrm{ave}}} = \frac{ \vec {\Delta s}}{\Delta t} [/tex]
      where [itex]\vec {\Delta s} [/itex] is the change in displacement and [itex] \Delta t [/itex] is the change in time (i.e., the specified time interval).

    [Edit: Instantaneous velocity and average velocity become equal when [itex] \Delta t \rightarrow 0 [/itex], at time t, where the instantaneous velocity was measured, meaning the two points on the curve merge into each other to form a single point. Otherwise, instantaneous and average velocities are not necessarily equal.]
     
    Last edited: Aug 4, 2013
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