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Homework Help: What does the area under a Distance vs Time graph indicate

  1. Aug 2, 2013 #1
    1. The problem statement, all variables and given/known data
    Given a graph with distance (in meters) on the y axis and time (in seconds) on the x axis what does the area under this graph illustrate ?

    The object has a velocity of 1m/s so after 5 s it has travelled 5m.

    2. Relevant equations

    This is a linear line (i.e y=x) so the area can be calculated based on that of a right angled triangle ( 1/2 base x height)

    3. The attempt at a solution

    So using the time of 5 seconds, we have a triangle area of 1/2(5s)* 5m = 12.5

    I'm unsure what these units would be ?? m/s ?? I didn't think it would be m/s as this is velocity and the velocity of the object is 1m/s.

    The area under graph , I don't believe to be acceleration, as to get acceleration from the area under a graph I'd need to have velocity on the y axis and time on the x axis....or wait would velocity on the y and time on the x give me displacement ???

    I have distance in m on the y axis and time in s on the x axis.

    I know I can calculate velocity at specific points on the graph but don't understand what the area under a distance vs time graph would show.
  2. jcsd
  3. Aug 2, 2013 #2
    You already figured out the units in your question.

    1/2(5s)*5m = 12.5 ms

    Apparently this is called absement. In ordinary kinematics I don't believe it has any physical or useful meaning. However there may be some theoretical use to the quantity in more advanced systems/problems.
  4. Aug 2, 2013 #3

    Simon Bridge

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    The graph is usually displacement-time rater than distance-time.

    The object starts with a displacement of zero and finishes with a displacement of 5m after 1s - is this correct?

    To find the units, use the same equation for the units that you used to get the number, but leave out all the bits that don't have units. In your case 12.5 m.s (since "ms"="milliseconds")

    The area under the displacement-time graph is the "absement" and does not have any special physical significance.
  5. Aug 2, 2013 #4
    Are you sure it's displacement - time graph or velocity - time graph, because the equation given 1/2bh is to get displacement under a velocity - time graph because area = displacement under the graph, and under a displacement - time graph really has no significance as mentioned by Simon.
  6. Aug 2, 2013 #5

    Simon Bridge

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    @oldspice: the problem statement describes an object moving at constant velocity ... wouldn't the (general geometric formula for the) area under the v-t graph for that be ##\small bh## rather than ##\small bh/2##?

    Note: having no special physical significance is not the same as having no significance. See examples:
    Last edited: Aug 2, 2013
  7. Aug 3, 2013 #6
    Well we don't really have the photo but for what we are given from op since its linear and a triangle geometrically we can only assume its 1/2bh otherwise I'd think he would have a trapezoid where it would then be 1/2b(h1+h2).
    Last edited: Aug 3, 2013
  8. Aug 3, 2013 #7

    Simon Bridge

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    That's right ... hence: considering it is a triangle and a constant velocity is involved ... then wouldn't that make it a d-t graph rather than a v-t graph? (Since a v-t graph for constant velocity would be a rectangle? re. post #4)
  9. Aug 3, 2013 #8
    Yes, a constant velocity - time graph will be a rectangle giving you displacement for the area. In his case we are dealing with a position - time graph, where there is absement as you mentioned earlier. If he wants velocity, he can just use the slope formula rise/ run. Another thing, an instantaneous velocity - time graph will have a triangle shape area that's where you would have to use 1/2bh to figure out the displacement.

    http://wearcam.org/absement/examples.htm for absement. (Simon mentioned this website as well)

    I'll say this is the first time I've seen a question like this though, odd.
    Last edited: Aug 3, 2013
  10. Aug 3, 2013 #9

    Simon Bridge

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    I don't normally see it in an assignment. The next question will be "what is the area under the absement-time graph?"

    I normally see this sort of extrapolation the other way: the slope of the v-t graph is the acceleration ... the slope of the a-t graph is the ... jerk ... the slope of the j-t graph would be the... what... putz? heel? prat? [*]

    But I suppose it makes sense ... in kinematics, the time derivative of the jerk is zero most everywhere terminating the series, but you can always integrate. For a constant acceleration, the d-t graph is quadratic, so the absement-t graph will be cubic etc. Perhaps the time-integral of the absement is a presement (since an absence of an absence is a presence?)


    [*] jounce :)
  11. Aug 4, 2013 #10
    Are you suggesting that 'putz', 'heel', and 'prat' are suitable alternatives for the gradient of the acc against time graph?
    Are you fully aware of the meaning of some of these words in some parts of the world?
  12. Aug 4, 2013 #11

    Simon Bridge

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    LOL - of course not :D
    Take another look
    ... the time-derivative of velocity is acceleration
    ... the time derivative of acceleration is jerk
    ... are you aware of the meaning of this last word in some parts of the World?

    ... continuing in this way, it seems logical that the time derivative of a jerk would be a heel and the time derivative of a heel would be a putz and so on. I met the "putz" version in a lecture on classical mechanics given by an American prof.

    It's a pune, or play on words.

    In tune with the multinational nature of the forums - "heel" would be mildly derogatory to Americans in general (possibly puzzling to others), "putz" would be similar with people familiar with Yiddish, and "prat" for UK/Commonwealth folk. All pretty mild.

    Of course, the time derivative of acceleration is actually called "jounce" ;)
    Mind you, if someone is seriously offended by any of this, just say the word and I'll happily report myself for moderation.
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