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Conceptual misunderstanding of Velocity-Time Graphs

  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.
     

    Attached Files:

  2. jcsd
  3. Apr 30, 2012 #2

    PhysicoRaj

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    The displacement is A-B. It is like this. In your graph the velocity rises to a certain extent and then comes constant. Then it drops down to zero. Now the body is not in motion. Since velocity is a vector, negative means the direction has reversed. So, as the line drops down you must understand that the body is reversing.Then absolutely the displacement will be less Than the amount it had covered before reversing. Hence the area/displacement would be A-B.
     
  4. Apr 30, 2012 #3
    but isnt the area under graph of B negative (since it is below x axis?)? if u were to use the values in the graph, area of B= 1/2 x 2 x -2 + 1 x -2 = -4. then A-B will become A-(-4) which is A + 4. That would be wrong right?
     
  5. Apr 30, 2012 #4

    PhysicoRaj

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    But see! The area of B (neg) is smaller than area of A (pos)
    A-B must give you a positive area.
    Logically acceptable......
     
  6. Apr 30, 2012 #5
    But you see if you did the working for the total displacement, you get:A = 1/2 x 2 x 3 + 2 x 3 + 1/2 x 3 x 3, B = (1/2 x -2 x 2) + (1 x -2)
    A + B gives you the actual displacement of 9.5m while A - B gives 17.5m, which is wrong
     
  7. Apr 30, 2012 #6
    isnt it?
     
  8. Apr 30, 2012 #7

    Curious3141

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    That's wrong.

    If you take the areas A and B as absolute values (which are always positive), then the *displacement* at 10s will be given as A - B and the total *distance* travelled at 10s is A + B.

    Areas are always positive, but you must be careful to distinguish the value of a definite integral from the area "under" a curve (or between a curve and the horizontal axis).

    In this case, [itex]\int_0^7v(t)dt > 0[/itex] whereas [itex]\int_7^{10}v(t)dt < 0[/itex]. The sum, which is equal to [itex]\int_0^{10}v(t)dt[/itex], which yields the displacement at 10s, is positive because the second integral (from 7 to 10) is of smaller magnitude than the first (from 0 to 7).

    The respective areas A and B are given by the absolute values of the integrals, i.e.

    [itex]A = |{\int_0^7v(t)dt}|[/itex] and [itex]B = |{\int_7^{10}v(t)dt}|[/itex] and the total distance = A + B.
     
  9. Apr 30, 2012 #8
    why must areas always be positive? i mean if you use the values from the graph area is negative right? and i havent learnt calculus yet (learning it next semester) so is there a simpler way of explanation please?
     
  10. Apr 30, 2012 #9

    HallsofIvy

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    The displacement (actual distance between the starting point and the ending point) is A- B. The distance traveled is A+ B. (If you to 90 m straight ahead the go back 30 m toward your starting point, you will end 90- 30= 60 m from the start but you will have traveled a total of 90+ 30= 120 m.)
     
  11. Apr 30, 2012 #10
    ok erm ya i get that but is area under graph always positive? since the B is below the x axis, B area should be negative right? if u took A = 90, B = -30 then wont the diplacement be A + B?
     
  12. Apr 30, 2012 #11

    Curious3141

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    Area is, by definition, positive. This link may help explain things simply: http://www.teacherschoice.com.au/maths_library/calculus/area_under_a_curve.htm

    I realise you haven't started calculus yet, but there's no escaping basic integration if you're talking about the area under a curve.
     
  13. Apr 30, 2012 #12
    In the link you provided, there was one part saying total shaded area = A1 + A2 + A3 even though A2 is below x axis. and it was written "but when the graph line is below the ‘x’ axis, the definite integral is negative. " doesnt that mean that area under graph can be negative when below x axis? Or is it just a misunderstanding of the statement by me?
     
  14. Apr 30, 2012 #13

    Curious3141

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    Misunderstanding by you.

    When the curve is below the x-axis, the definite integral taken in the direction of increasing x (with the higher x value as the upper bound) will be negative. In that case, the area is specifically defined as MINUS (one) times the integral, so that the area remains positive (minus times minus = plus). Note carefully how they defined A2, and how it differs from how they defined A1 and A3.

    So the area is always positive. I don't know where you're getting the idea that area can ever be negative.
     
  15. Apr 30, 2012 #14
    Thanks for all the help into making me understand why area must be positive!!
     
    Last edited: Apr 30, 2012
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