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Describing the motion of a graph.

  1. Feb 18, 2012 #1
    Hi i have a graph given in kinematics. I need to describe the motion and calculate the average velocity during the entire duration. I have attached a picture. How would i calculate the average velocity

    No equations give

    I said that the velocity is not constant. I dont know if the average velocity woul dbe negative?

    Attached Files:

  2. jcsd
  3. Feb 18, 2012 #2
    How would you define average velocity?

    [itex]\Delta v = \frac{\Delta x}{\Delta t}[/itex]

    Something like that, yes?
    Use that definition, what comes out?
  4. Feb 18, 2012 #3
    Thanks for your reply. But would the average velocity be negative looking at the trend of the graph? And how would i describe this motion?
  5. Feb 18, 2012 #4


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    From your graph, it looks like x is 25 when t is about 2 and x is about 2 when t= 30. So what are [itex]\Delta x[/itex], [itex]\Delta t[/itex], and [itex]\Delta x/\Delta t[/itex]?
  6. Feb 18, 2012 #5
    If you apply the definition of the average velocity you will see that it is negative since the change in position is definitely negative (it seems to start at some high x and ends up at x=0)
    So yes, the average velocity of would be negative.

    To describe the motion, we could look at the average velocity between different times and see how they relate.

    Between t=10 and t=20 the position has gone from x=15 to x=5, so the average velocity during this time is given as'

    [itex]\Delta v = \frac{\Delta x}{\Delta t} = \frac{-10}{10} = -1[/itex]

    If we look at the average velocity at some later time, say t=20 and t=30 the position has gone from x=5 to perhaps x~2 and so

    [itex]\Delta v = \frac{\Delta x}{\Delta t} = \frac{-3}{10} = -0.3[/itex]

    So we can see that the average velocity is decreasing as time increases.
    We can also see that the position is finally going to reach x=0 as t gets to be quite big.

    We could describe this by saying that the position 'asymptotes' to x=0 for large t and that the velocity also 'asymptotes' to 0 for large t.

    Do you understand my reasoning here?
  7. Feb 18, 2012 #6
    Yes but what is the general description of this motion. Like is the object moving in an westward direction?
  8. Feb 18, 2012 #7
    That depends on what direction you want to set as the positive x direction.
    If you set west as your positive x direction then you're moving westward, if you set north as your positive x direction then you're moving northward.
    It is completely up to you which direction you set positive x to be facing
  9. Feb 18, 2012 #8
    I would say that approximatey 25m was covered in approximately 30 to 40 secs.
    So the only sensible answer for me is that the average velocity is between 25/30 and 25/40 m/s
    ie 0.83m/s and 0.63m/s
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