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Test question

  1. Sep 20, 2011 #1
    The question was, Explain instantaneous velocity.

    My answer:
    Limit as delta(t) approaches zero of average velocity is inst. velocity. Or as delta(t) approaches zero, average velocity approaches inst. velocity.
    inst. velocity is the slope of the tangent line at a given point.....

    Then I wrote:
    V = Lim Delta(Average Velocity)/Delta(t)
    Delta(t) --> 0


    Is my answer totally wrong because I got 5/10 pts on the answer?
     
  2. jcsd
  3. Sep 20, 2011 #2

    NascentOxygen

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    Staff: Mentor

    tangent (of what graph) gives magnitude. velocity also needs a direction.
     
  4. Sep 20, 2011 #3
    This is the picture I added....
     

    Attached Files:

  5. Sep 20, 2011 #4
    Isnt saying slope of a tangent line at a "given point" general? It can be any point......
     
  6. Sep 20, 2011 #5

    Hootenanny

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    Staff Emeritus
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    Gold Member

    Your answer isn't totally wrong - its only half wrong, that's why you got half marks :wink:.

    The first thing to point out is that the average velocity is

    [tex]v_\text{ave} = \frac{\Delta x}{\Delta t},[/tex]

    so your limit becomes

    [tex]V = \lim_{\Delta t\to0} \left(\frac{\Delta x}{\Delta t}\right)/\Delta t = \lim_{\Delta t\to0} \Delta x,[/tex]

    which obviously isn't correct. So the limit you wrote, wasn't correct. The first part however, was okay:
    However, your next sentence wasn't very clear:
    The tangent line to what? What are you plotting?

    A better answer would have been as follows. The instantaneous velocity is equal to the average velocity in the limit as [itex]\Delta t\to0[/itex], or more generally, the derivative of position with respect to time. The velocity at a given point in time is equal to the gradient of the displacement at that point in time on a displacement-time graph.

    In one dimension,

    [tex]v = \lim_{t\to t_0} \frac{x(t_0)-x(t)}{t-t_0} = \frac{\text{d}x}{\text{d}t}.[/tex]

    Does that make sense?
     
    Last edited: Sep 20, 2011
  7. Sep 20, 2011 #6

    NascentOxygen

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    Staff: Mentor

    Well, there's your mistake. That slope is acceleration! :cry:

    An "average velocity" graph doesn't make much sense (in this context). A velocity graph is velocity versus time, and that's instantaneous velocity, anyway. Or speed.

    You were thinking of a position versus time graph, and finding the tangent to it.
     
    Last edited: Sep 20, 2011
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