Area under the velocity curve

  • Thread starter madah12
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  • #1
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Is the area under the velocity - time curve the displacement or the distance? I think it is the distance because areas are not affected by negative velocities.
Edit:
in the book Haliday and Resnick Fundamentals of physics 8th edition page 27
"When the acceleration curve is above the
time axis, the area is positive; when the curve is below the time axis, the area is
negative." how is that right ? I mean areas are never negative right?
 
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Answers and Replies

  • #2
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It is the displacement, area that is below the x-axis is negative.
 
  • #3
326
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yes but in the area when the integral is negative we subtract to get a positive area right? I mean if the integral from 1 to 3 is positive and from to 3 to 4 is negative we subtract them so we dont get the effects of the negative.
Edit:
In my highschool in calculus when we are asked to find the area between the function and the x axis we find the roots of the function and separate the integral into intervals where we add in incase of positive intervals and subtract incase of negative ones.
 
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  • #4
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yes but in the area when the integral is negative we subtract to get a positive area right? I mean if the integral from 1 to 3 is positive and from to 3 to 4 is negative we subtract them so we dont get the effects of the negative.

That's exactly why you DO get the effects of the negative. The stuff below the x-axis is subtracted, not added--so you are including the effects. If the curve was ENTIRELY below the axis, you would get a negative answer. If Half was on top, and half below (by area) you get 0 (displacement), instead of 2*top-half (distance).

And the integral automatically takes the top/bottom into account. You don't have to split it up. I.e. the information is included in the function.
 
  • #5
326
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ok wait I am not an expert but I am saying that we have a graph like x^2 -4x -4 with roots of 2( 1-sqrt(2)) and 2(1+sqrt(2)) when we calculate the area from -3 to 3 we do it like this integral from -3 to 2(1-sqrt(2)) f(x) dx (which is positive because the function is above the axis) - integral from 2(1-sqrt(2)) to 2(1+sqrt(2)) f (x) dx (which is negative so when we subtract it we negative of negative is positive) + integral from 2(1+sqrt(2)) to 3 f(x)dx(which is positive) so how can an area be negative? or do you do it in another way.
 
  • #6
ehild
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No, we do not subtract the negative area, we add them together, either positive or negative. It is the same as adding up positive and negative numbers.

The integral is not equivalent with the geometric area a curve encloses with the horizontal axis.

ehild
 
  • #7
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No, we do not subtract the negative area, we add them together, either positive or negative. It is the same as adding up positive and negative numbers.

The integral is not equivalent with the geometric area a curve encloses with the horizontal axis.

ehild

I know that in integrals we don't but when we find areas we do I am saying if we find the area under the v-t curve and treated the negative area as positive and not subtracted it do we get the distance?
 
  • #8
ehild
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I know that in integrals we don't but when we find areas we do I am saying if we find the area under the v-t curve and treated the negative area as positive and not subtracted it do we get the distance?

No. You need to treat is negative and add or positive and subtract.

ehild
 
  • #9
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ok what about if we take the integral of from t1 to t2|v|dt isnt that the distance?
 
  • #10
ehild
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It is the distance travelled, and it is positive because of the absolute value.
If you walk forward and then backwards with the same speed v and time t, your displacement will be vt+(-vt)=0, but the distance travelled is 2|v|t.

ehild
 

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