Physics Graph Problem

[Godric]

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I was using ExamKrackers 1001 physics and doing a problem that states

#72. "The total area between the curve and the zero time axis on any velocity versus time graph represents:"

A. Distance <----- Book answer
B. Displacement <---- My answer
C. Velocity
D. Acceleration

I remembered from Chads Physics video on graphs, that the area under velocity vs time = displacement, and I went to check the answer in the back and it states that it is Distance not Displacement, "The total area would assume that the area beneath the x axis is positive. This indicates distance not displacement".
So my question is EK1001 right and did I misunderstand the whole concept or is EK wrong ?

 

[milski]

They are correct. Technically, the area is always a positive, which means that it will represent and not distance. The two are the same if the object is moving only in one direction but if the object starts moving back, the displacement will start to decrease while the distance will continue to increase.

 

[MedPR]

I was using ExamKrackers 1001 physics and doing a problem that states

#72. "The total area between the curve and the zero time axis on any velocity versus time graph represents:"

A. Distance <----- Book answer
B. Displacement <---- My answer
C. Velocity
D. Acceleration

I remembered from Chads Physics video on graphs, that the area under velocity vs time = displacement, and I went to check the answer in the back and it states that it is Distance not Displacement, "The total area would assume that the area beneath the x axis is positive. This indicates distance not displacement".
So my question is EK1001 right and did I misunderstand the whole concept or is EK wrong ?

The book is right.

If a specific curve is only in quadrant 1, then the area under that curve is displacement since the velocity was never negative (in the opposite direction). If a v/t graph dips into the negative v direction (below the x axis) then you know that the thing was moving in the negative (opposite) direction, thus reducing the displacement but continuing to add to the distance.

 

[Godric]

The book is right.

If a specific curve is only in quadrant 1, then the area under that curve is displacement since the velocity was never negative (in the opposite direction). If a v/t graph dips into the negative v direction (below the x axis) then you know that the thing was moving in the negative (opposite) direction, thus reducing the displacement but continuing to add to the distance.

So as long as I am being shown q1 and nothing goes into q4 (neg v) then the area under the curve is always equal to displacement?

And if thats the case then the answer doesnt make sense, because the graphs are all in q1.

 

[milski]

If the graph is only in q1, displacement and distance will have the same value. They are still two distinct concepts. I guess that's what they're trying to test with this question. No idea if the real MCAT is that picky.

 

[Godric]

If the graph is only in q1, displacement and distance will have the same value. They are still two distinct concepts. I guess that's what they're trying to test with this question. No idea if the real MCAT is that picky.

Thank you, I suppose it was a badly worded question also because the question says "on any graph" and yet the directions say refer to the graphs that they have provided. But I do understand the concept better than I did before.

 

[MedPR]

So as long as I am being shown q1 and nothing goes into q4 (neg v) then the area under the curve is always equal to displacement?

And if thats the case then the answer doesnt make sense, because the graphs are all in q1.

Yes and no.

Yes if the question specifically refers to those 4 graphs only. So yes, for all 4 of those graphs, the area between the curve and the x axis is the displacement.


No because it doesn't really matter what graphs or data they are showing you unless they are referring to it. The question was

#72. "The total area between the curve and the zero time axis on any velocity versus time graph represents:"

. The key phrase is "any velocity versus time graph...". This implies v/t graphs including, but not limited to the ones presented to you.

If I move at 10m/s for 10 seconds, then I move at -20m/s for 10 seconds, what distance did I travel? What is my displacement? What is the area under that curve? Draw that out, and answer all 3 questions, then the answer/explanation will be clear.

 

[milski]

Yes if the question specifically refers to those 4 graphs only. So yes, for all 4 of those graphs, the area between the curve and the x axis is the displacement.

I would not phrase it that way, it will only lead to trouble. Even in these graphs the area is still the distance. It just happens that its value is the same as the displacement. That might seem like splitting hairs but I think it should be less confusing, long term.