omc1
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Homework Statement
A plot of Calvin's Mom's position vs. time is shown. Rank the acceleration of the five regions (e.g., A, B, etc.) from most negative to most positive
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How Do You Analyze Acceleration on a Position vs. Time Graph?
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A plot of Calvin's Mom's position vs. time is shown. Rank the acceleration of the five regions (e.g., A, B, etc.) from most negative to most positive
See attachment!
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omc1 said:
This is position vs. time, so the derivative (slope at any point) of this graph gives the velocity.
The rate of change of the velocity is the acceleration.
So, on this graph:
Where is the slope increasing?
Where is the slope decreasing?
Where is the slope constant?
Where is the slope decreasing?
Where is the slope constant?
omc1
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well, the slope is increasing at a, and maybe at e, its constant at c, and decreasing at d, then at b it is slowing down
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Is the slope (not the function itself) really decreasing in D?
The other parts are right.
The other parts are right.
omc1
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well at first i thought d was constant and since slope relates to the velocity then a is zero but iam just really confused when trying to read this graph
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Your first thought was right. D corresponds to a constant, negative velocity. No acceleration there.
omc1
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ok that makes sense, thanks!
th3c00n
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Bumping an old thread for a little help on a similar question.
I am given a position vs time graph:
I am asked where acceleration is positive and where it is negative.
I believe I understand that when the slope of the line is increasing then the acceleration is positive. When the slope of the line is decreasing then the acceleration should be negative. Correct?
So from what I see here acceleration at the given points should be:
A - Negative
B - 0
C - Negative
D - Positive
E - 0
F - Positive
Yet these answers are being marked as incorrect when I submit them. Can PF lend me a hand here? Thank you!
I am given a position vs time graph:
I am asked where acceleration is positive and where it is negative.
I believe I understand that when the slope of the line is increasing then the acceleration is positive. When the slope of the line is decreasing then the acceleration should be negative. Correct?
So from what I see here acceleration at the given points should be:
A - Negative
B - 0
C - Negative
D - Positive
E - 0
F - Positive
Yet these answers are being marked as incorrect when I submit them. Can PF lend me a hand here? Thank you!
Toranc3
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th3c00n said:Bumping an old thread for a little help on a similar question.
I am given a position vs time graph:
I am asked where acceleration is positive and where it is negative.
I believe I understand that when the slope of the line is increasing then the acceleration is positive. When the slope of the line is decreasing then the acceleration should be negative. Correct?
So from what I see here acceleration at the given points should be:
A - Negative
B - 0
C - Negative
D - Positive
E - 0
F - Positive
Yet these answers are being marked as incorrect when I submit them. Can PF lend me a hand here? Thank you!
I think B and E are incorrect.
Curvature upward acceleration is positive and curvature downward acceleration is negative.
th3c00n
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B and E appear to be where the object changes direction. An earlier part of the question asked where velocity was 0 and those were the correct answers. Can you have acceleration without velocity?
th3c00n
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Apparently you can, that was the correct answer, thank you!
Toranc3
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th3c00n said:
Well I thought that you were looking for acceleration.
Toranc3
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th3c00n said:
Yo if the graph is position vs time then velocity would be zero when the slope is zero but if they ask for acceleration then its different. (refer back to my original post).
th3c00n
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Toranc3 said:
Yep, got it now. I guess I wasn't thinking of local min/max points as a change in slope. Not sure why... Ha. Thank you again!
Toranc3
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th3c00n said:
Yeah no problem! Can you help me on my post? :)
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I see that you're new to PF, so Welcome!th3c00n said:Bumping an old thread for a little help on a similar question.
I am given a position vs time graph:
...
I should point out that according to the rules of this forum, you should have started a new thread to post this rather than "highjacking" an existing thread.