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Conceptual Area Under Velocity-Time graph confusion

  1. Apr 30, 2012 #1
    Hi all I have a confusion regarding a physics concept. When you consider area under graph (velocity time graph), if the area is below x axis, is the area considered to be negative?
    For example, look at the attachment. There are 2 shaded areas. Let the area above x axis be A and area below x axis be B and ignore all the numerical values on the graph. Is the displacement A+B or A-B? I am very confused.
     
    Last edited: Apr 30, 2012
  2. jcsd
  3. Apr 30, 2012 #2

    sophiecentaur

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    I can't see your attachment but I can say that you can treat areas under the x axis as negative (after all, the velocity IS negative, isn't it? - so a positive time times a negative velocity gives you a negative answer.). So A+(a negative)B is the total area.
    Alternatively, and less strictly mathematically, you can just subtract the areas underneath from the areas on top. Same answer, of course.
     
  4. Apr 30, 2012 #3
    The displacement is A-B, the distance is A+B.
     
  5. Apr 30, 2012 #4

    sophiecentaur

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    Only if you neglect that areas below the x axis are negative.
     
  6. Apr 30, 2012 #5
    I don't agree, areas are always positive or zero.
     
  7. Apr 30, 2012 #6

    sophiecentaur

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    I guess it's only a matter of taste but x times y will be negative if y is negative, surely?

    If you were using a computer to calculate displacement from a set of v,t coordinates, would you have a special routine to spot negative velocity sections and then 'subtract' them? I think you would, more likely, just add em all up with regard to the sign of the v.

    Would you also think of having 'acceleration' and 'deceleration', rather than positive and negative acceleration? There is a direct parallel there.
     
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