Very trick velocity vs. time graph

AI Thread Summary
The discussion revolves around understanding a complex velocity vs. time graph derived from a variable acceleration graph in a force and motion lab. Participants emphasize the importance of recognizing that positive acceleration leads to increasing velocity, even when the acceleration is decreasing. The confusion arises from visualizing sharp transitions in the graph, which should instead be smooth curves. One user notes that while the acceleration graph slopes downward, it remains positive, indicating that velocity continues to increase, albeit at a slower rate. Overall, the conversation highlights the challenge of interpreting these graphs but reassures that with practice, visualization becomes easier.
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Very tricky velocity vs. time graph

Homework Statement


Doing a very basic lab on force and motion and ran into a snag with a graph. It's a very wildly changing acceleration graph, and I'm unsure how to proceed with the matching velocity graph. I know that a curved acceleration graph would indicate an even more steeply curved velocity graph, but I'm mentally visualizing too many positive and negative accelerations to get my head around this. Please help.

Homework Equations



http://usera.imagecave.com/hotrod73dart/question/Vgraph.jpg

The Attempt at a Solution



http://usera.imagecave.com/hotrod73dart/question/vgraphattempt1.jpg
 
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Have you taken a calculus class yet? If you have not those sharp points in your velocity graph (i.e. at t = 1) do not have continuous derivatives, so that cannot be the answer because, well, you have the acceleration graph which is continuous.

Remember that a = dv/dt so it's the change in the velocity. So where ever the acceleration is positive the velocity will be increasing, does this help at all?
 
Feldoh said:
Have you taken a calculus class yet? If you have not those sharp points in your velocity graph (i.e. at t = 1) do not have continuous derivatives, so that cannot be the answer because, well, you have the acceleration graph which is continuous.

Remember that a = dv/dt so it's the change in the velocity. So where ever the acceleration is positive the velocity will be increasing, does this help at all?

No calculus yet unfortunately. Those sharp points are supposed to be sharply rounded curves, but I don't think that makes much sense either. I'm aware of what acceleration is, but when acceleration is changing at a non-constant rate (curved acceleration graph), then that means that velocity is increasingly increasingly increasing (or increasingly increasingly decreasing). Not an easy thing to visualize. Add positive and negative acceleration and positive and negative velocity into the mix and it gets really confusing.
 
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Maybe this will help: http://filer.case.edu/pal25/pic.jpg

You're right it is hard to visualize the first time you see it, but it gets easier the more you analyze these graphs.
 
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Feldoh said:
Maybe this will help: http://filer.case.edu/pal25/pic.jpg

You're right it is hard to visualize the first time you see it, but it gets easier the more you analyze these graphs.

If my newest attempt is correct then yes, your image helped very much. Otherwise, I'm still lost.

http://usera.imagecave.com/hotrod73dart/question/vgraphattempt3.jpg
 
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Yes, that looks right.
 
Looks good...the only thing is that the final velocity might not be that high.

As Feldoh said, it's hard to visualize, but the more experience you get, the easier it gets.
 
Feldoh said:
Yes, that looks right.

Wellesley said:
Looks good...the only thing is that the final velocity might not be that high.

As Feldoh said, it's hard to visualize, but the more experience you get, the easier it gets.

Thanks so much. It was very difficult until Feldoh pointed out the fact that while the acceleration graph was sloping downward it was still a positive value, meaning that velocity was still increasing, just not as largely incremental. Then it was easy to extrapolate that the negative acceleration merely negated the initial positive acceleration and brought the velocity back to the original value.

I love physics, but sometimes I feel it's going to be the death of me (and this is only 100 level!)
 
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