How do you draw an acceleration time graph?

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GogetaORvegito?
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Is the image attached the correct way where you just have to draw dotted lines connecting the different accelerations together
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Rhyall
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(Original post by GogetaORvegito?)
Is the image attached the correct way where you just have to draw dotted lines connecting the different accelerations together
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GogetaORvegito?
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(Original post by Rhyall)
So you don't need to connect any lines to anything?
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Rhyall
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(Original post by GogetaORvegito?)
So you don't need to connect any lines to anything?
Take the advice from the guy in the video, I have no clue, therefore I cannot answer this question.
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Eimmanuel
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(Original post by GogetaORvegito?)
Is the image attached the correct way where you just have to draw dotted lines connecting the different accelerations together
The unit of acceleration in the image is incorrect. Without the velocity vs time graph, we cannot tell whether the graph is correct or wrong.

(Original post by GogetaORvegito?)
So you don't need to connect any lines to anything?
Based on the velocity vs time graph in the video, there is no well-defined gradient the time where the acceleration changes, so we cannot connect the acceleration by solid lines. This implies that we don't connect the acceleration.

Similarly, for your graph shown in post #1, I believe there is no well-defined gradient in the velocity-time graph at t = 0.5s, 1.0s and 1.5s, so we connect the acceleration using dotted lines instead of solid lines.
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GogetaORvegito?
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(Original post by Eimmanuel)
The unit of acceleration in the image is incorrect. Without the velocity vs time graph, we cannot tell whether the graph is correct or wrong.



Based on the velocity vs time graph in the video, there is no well-defined gradient the time where the acceleration changes, so we cannot connect the acceleration by solid lines. This implies that we don't connect the acceleration.

Similarly, for your graph shown in post #1, I believe there is no well-defined gradient in the velocity-time graph at t = 0.5s, 1.0s and 1.5s, so we connect the acceleration using dotted lines instead of solid lines.
What do you mean by "not well defined in the video". Isn't the acceleratin the gradient of each of the different velocities that are shown. Also, I didn't know anything about that dotted line stuff so thanks. I initially thought you connect the lines to the x axis
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OxMus
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(Original post by GogetaORvegito?)
What do you mean by "not well defined in the video". Isn't the acceleratin the gradient of each of the different velocities that are shown. Also, I didn't know anything about that dotted line stuff so thanks. I initially thought you connect the lines to the x axis
It seems to me that the graph incorporates the assumption that the acceleration changes instantaneously, and that’s why the lines jump between a-values. This wouldn’t happen in real life, but that’s ok, because mechanics always uses some simplifying modelling assumptions to make the maths work.

At t = 0.5, 1, and 1.5, it doesn’t show bold vertical lines because that would suggest an undefined value for acceleration at those t-values (think: at t = 0.5, 1, or 1.5, what are the values of a? ... ???).

The graph isn’t quite correct because it ought to show open and closed circles enclosing the horizontal lines: closed means ‘equal to’ and open means ‘not equal to’. E.g. if there’s an open circle at (1.0,-0.4) and a closed circle at (1.0,1.2), then at t = 1.0, a = 1.2. This rectifies the problem of undefined a-values outlined above.

I hope this helps.

OxMus.
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GogetaORvegito?
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(Original post by OxMus)
It seems to me that the graph incorporates the assumption that the acceleration changes instantaneously, and that’s why the lines jump between a-values. This wouldn’t happen in real life, but that’s ok, because mechanics always uses some simplifying modelling assumptions to make the maths work.

At t = 0.5, 1, and 1.5, it doesn’t show bold vertical lines because that would suggest an undefined value for acceleration at those t-values (think: at t = 0.5, 1, or 1.5, what are the values of a? ... ???).

The graph isn’t quite correct because it ought to show open and closed circles enclosing the horizontal lines: closed means ‘equal to’ and open means ‘not equal to’. E.g. if there’s an open circle at (1.0,-0.4) and a closed circle at (1.0,1.2), then at t = 1.0, a = 1.2. This rectifies the problem of undefined a-values outlined above.

I hope this helps.

OxMus.
Oh ok then thanks for that
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Eimmanuel
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(Original post by GogetaORvegito?)
What do you mean by "not well defined in the video". Isn't the acceleration the gradient of each of the different velocities that are shown. Also, I didn't know anything about that dotted line stuff so thanks. I initially thought you connect the lines to the x axis
If you consider the modulus function of  f(x) = |x| , at x = 0, there are “many” tangent lines as discussed in the following thread in math stackexchange.
https://math.stackexchange.com/quest...value-function
The “many” tangent lines at x = 0 implies that there is no well-defined gradient at x = 0.


A well-defined gradient means there is only one and one tangent line at that point.
When there are “bends” in the velocity-time graph, the gradients at the “bends” are not well-defined.
“Bends” are like the point at t = 5s and 10 s as shown in the video that displays a velocity-time graph.


Adding to what OxMus had explained, (IMO) based on the velocity-time graph without any actual maths functions that describe the graph, there is no need to place an open or a close circle in the acceleration-time graph.

The practice of placing an open or close circle in an acceleration-time graph is usually ignored at A level physics. However, at the university level, your instructors or lecturers would provide the advice based on the course syllabus or requirements.
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