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Displacement as Function of Time Graph Question

  1. Sep 14, 2013 #1
    1. The problem statement, all variables and given/known data
    The graph below shows the vertical displacement of an object as a function of time.
    (a) Construct a graph of the object’s velocity as a function of time.
    (b) Construct a graph of the object’s acceleration as a function of time.
    (c) Describe a scenario in which a real-life object might move in this way


    3. The attempt at a solution

    Ok, Here is a picture. Graph A represents the original graph that we construct B and C from. I am not sure if I drew graph B (velocity as a function of time) and C (acceleration as a function of time) correctly. Did I?
    15f4y6r.png

    Also for a potential scenario, I can't come up with anything except for maybe a helicopter that goes up to a certain height, then hovers at that height then comes back down. But I'm not sure about the curvatures on that original graph A so I want to make the scenario detailed.

    Thanks!
     
  2. jcsd
  3. Sep 15, 2013 #2
    Before I say anything useful I will say that it has been a very long time since I've looked at position (or velocity or acceleration) vs. time graphs so it's possible that I am wrong but: I think your graphs are fine (but you shouldn't trust me entirely).

    One thing I will mention is that you should aim to provide titles and label your axes for ease of other readers (ie. us "helpers" or your teacher). Some (many?) teachers will be a stickler for such things and it will be easy to lose marks in a silly way if you don't label your axes carefully.

    As for a scenario, I believe the one you have described works well. We could try to be slightly more detailed. On the position vs. time graph, prior to peaking we notice that the slope decreases slightly - this means that the velocity is somewhat decreased.

    We can explain this in the following way: A helicopter lifts up off of the ground and moves upwards with some speed. At a certain height the pilot decides to slow his ascent and hover in the air at some height. Then he descends in a similar fashion. (This is basically what you said).

    We could make this more elaborate.

    Suppose we are on a planet with trees that grow very, very tall. One day, your kitten, Paws, decides to climb one of these trees (she is chasing a delicious looking bird-like creature) and gets stuck very high up. You contact the emergency services and the send a rescue helicopter to save her. The pilot flies his helicopter up and as he approaches the location of Paws he begins to slow down and hover in mid-air when he is near. The passenger, a professional cat rescuer rescues Paws and brings her on board the helicopter and the pilot begins the descent again.
     
  4. Sep 15, 2013 #3

    NascentOxygen

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    Hi physicsnobrain. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    Take a ruler and lay it to be tangential at some point on your plot of distance vs time, the steepness of the incline of the ruler here indicates the velocity of the moving object at that moment. The steeper the slope of ruler, the greater the speed.

    Slowly move the ruler along the curve, adjusting it so it stays tangential to the curve, and you get a clear picture of how the speed of the object is changing (or not changing, as the case may be). When the ruler's incline is steep, the speed is greater.

    Ignore the graphs you have already sketched. Start anew and work out bit by bit how the velocity vs time graph should go.
     
    Last edited by a moderator: May 6, 2017
  5. Sep 15, 2013 #4
    Are the graphs I drew not correct? Also, since the velocity time is velocity as a function of time, if I drew it steep that would mean there is acceleration, which I believe there isn't judging from the graph A.
     
    Last edited by a moderator: May 6, 2017
  6. Sep 15, 2013 #5
    Ok, after taking your advice this is now the graph I get of velocity as a function of time constructed from the graph A in my first post.
    34g7e9w.png

    Is it right?
     
    Last edited by a moderator: May 6, 2017
  7. Sep 15, 2013 #6
    And here is the acceleration as a function of time graph.
    msoh07.png

    Again, is this right?
     
  8. Sep 15, 2013 #7
    Does anybody have an idea if my two new graphs work given graph A in my original post?
     
  9. Sep 15, 2013 #8

    NascentOxygen

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    That is looking better. Though we don't know that those sloping straight lines should actually be straight lines, it is better than your first post's graph (B).

    The right hand side of your plot should not end like you show it, though.

    EDITED
     
    Last edited: Sep 15, 2013
  10. Sep 15, 2013 #9
    Is this better for velocity as function of time?
    2i77gw1.png
     
  11. Sep 15, 2013 #10
    Also, would the acceleration just be a straight line at zero?
     
  12. Sep 15, 2013 #11

    NascentOxygen

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    No, that is not better. A period of constant velocity would show on the displacement-time plot as a straight (though sloping) line. There are no such lines on the distance-time plot given for this question. What you had was good enough for now.

    You have noticed how the distance-time graph ends with the object sitting at zero distance for some time?

    Only when you have the velocity-time plot right is it worth discussing the acceleration.
     
  13. Sep 15, 2013 #12
    2mq6p88.png

    Ok I drew this velocity as a function of time graph. This is accurate, correct?
     
  14. Sep 16, 2013 #13

    NascentOxygen

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    Pay attention to two regions:

    attachment.php?attachmentid=61863.jpg
     

    Attached Files:

  15. Sep 16, 2013 #14
    That second region you circled shouldnt be there. It was a mistake on my part when i drew the graph in paint. Pretend that the graph ends when it hits the x axis.
     
  16. Sep 16, 2013 #15

    NascentOxygen

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    Okay. But how did you decide (on the right hand side of the plot) there is a sudden bend in the line halfway while the object returns to zero level?

    The arrow points to a point in time where distance was rapidly changing with time then it abruptly stopped changing. That happens when speed has an abrupt drop to zero.
     
  17. Sep 16, 2013 #16
    Ok, I see the abrupt change. I went back to the graph I drew before and modified the second part.

    oj42eb.png

    Is this more like it?
     
  18. Sep 16, 2013 #17

    haruspex

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    You have the first abrupt change to zero velocity correct, but you are still not showing the second one. When the distance graph finally returns to the t axis, it stays there, so velocity is now zero.
    Also, because distance ends up where it started, at zero, the total area under the velocity graph (counting areas below the t axis as negative) should be zero. I.e. the total area above the t axis should be equal to the total area below it.
     
  19. Sep 16, 2013 #18

    NascentOxygen

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    The left hand half of the plot is looking good. Next, that 120 degree bend you show in the right hand half, why did you draw it like that?
     
  20. Sep 16, 2013 #19
    Well because there is no abrupt change in the second half of the graph, it flows.
     
  21. Sep 16, 2013 #20

    NascentOxygen

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    So why did you show an abrupt change in the way velocity changes where there is no abrupt change in the way displacement is changing?

    Anyway, you should get back to the graph you drew here. https://www.physicsforums.com/showpost.php?p=4503814&postcount=5

    34g7e9w.png

    Now that you cleared up the artifact I had circled, that graph from earlier is correct!

    Now you can determine the acceleration vs time graph. (The one in your very first post is not correct.)
     
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