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Instant and average velocity

  1. Nov 7, 2005 #1
    if I was given a graph, and told to work out the instant and average velocity, do I just find the slope of a tangent for both instant and average velocity? thx~
     
  2. jcsd
  3. Nov 7, 2005 #2
    Instant velocity requires finding the tangent while average velocity is [itex]v_{avg} = \frac{\Delta x}{\Delta t}[/itex].
     
  4. Nov 7, 2005 #3
    Just expanding on what cscott said,

    **Average velocity = [tex] {\Delta x / \Delta t} [/tex], --->i.e., displacement (change in position) over the change in time. Remember that [tex] \Delta x[/tex] here refers to displacement, not the distance traveled.

    **Instantaneous velocity is the instantaneous rate of change of position. We can calculate instaneous velocity via the first derivative of position with respect to time. As you can see,
    Instantaneous velocity = [tex] {dx/dt} [/tex].
     
  5. Nov 7, 2005 #4
    um... so say if were to find the instant velocity I would basically just draw a line in the grid through the one point of the graph, then find the slope of that line, and that'll give me the instant velocity of that point. And if I were to find the average velocity between two points (displacement) I would find the slope of the line that goes from one point to the other right?

    And also, could a velocity-time graph show the displacement? There is a question about how a car goes 70km and 20km in the reverse direction. I know that the displacement from the starting point would be 50 but could that be shown on a velocity-time graph?
     
  6. Nov 7, 2005 #5
    Yes. The area under the curve of where you started to where you ended would give you displacement, but not total distance. If you're taking Calculus, you can recognize the fact that

    [tex]\int_{t1}^{t2} \vec{v}dt=S[/tex].
     
  7. Nov 7, 2005 #6
    okay~ thank you! ^_^
     
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