Conceptual Area Under Velocity-Time graph confusion

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Discussion Overview

The discussion revolves around the interpretation of the area under a velocity-time graph, particularly concerning whether areas below the x-axis should be considered negative and how this affects calculations of displacement and distance. The scope includes conceptual understanding and mathematical reasoning related to physics principles.

Discussion Character

  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions if the area below the x-axis in a velocity-time graph should be considered negative, asking if displacement should be calculated as A+B or A-B.
  • Another participant suggests that areas under the x-axis can be treated as negative, leading to a total area calculation of A+(a negative)B.
  • Some participants assert that displacement should be calculated as A-B and distance as A+B, but note that this assumes areas below the x-axis are treated as negative.
  • Contrarily, other participants argue that areas should always be considered positive or zero, questioning the necessity of subtracting areas below the x-axis.
  • One participant raises a point about the practical approach of calculating displacement from velocity-time coordinates, suggesting that negative velocities would naturally lead to negative contributions without needing special routines.
  • There is a discussion about the terminology of acceleration and deceleration, drawing parallels to the treatment of negative areas in the graph.

Areas of Agreement / Disagreement

Participants express differing views on whether areas below the x-axis should be treated as negative, leading to a lack of consensus on how to calculate displacement and distance from the graph.

Contextual Notes

Some participants' arguments depend on the interpretation of negative areas and the definitions of displacement and distance, which remain unresolved in the discussion.

zabachi
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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.
 
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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.
 
The displacement is A-B, the distance is A+B.
 
MikeyW said:
The displacement is A-B, the distance is A+B.

Only if you neglect that areas below the x-axis are negative.
 
I don't agree, areas are always positive or zero.
 
MikeyW said:
I don't agree, areas are always positive or zero.

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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