Displacement as Function of Time Graph Question

In summary, After receiving feedback and advice, the velocity and acceleration graphs were redrawn based on the original graph A. The velocity graph shows a decrease in slope before peaking, indicating a decrease in speed before hovering. The acceleration graph shows a slight positive slope throughout, indicating a constant acceleration. These graphs could represent a scenario where a helicopter hovers at a certain height before descending. However, labeling of axes and titles should be included for accuracy.
  • #36
physicsnobrain said:
Ok. But at the end it accelerates slow then quickly accelerates. So should it be this?
3536h.png
No. It should essentially be exactly the same as #12, but with a vertical line coming up from the final point at the right-side, up to the t axis.
 
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  • #37
Chestermiller said:
No. It should essentially be exactly the same as #12, but with a vertical line coming up from the final point at the right-side, up to the t axis.

Could you please explain why it makes that bend in the right hand side of the graph?
 
  • #38
physicsnobrain said:
Ok, well then this has to be the right answer.
You are allowed to say that only once! :smile:

etxjzd.png

Yes, that's about as correct as you are going to manage here. It should sync with the distance vs time graph, so that corresponding times line up. The original distance vs time graph has 3 intervals of roughly equal duration. (That final vertical line is debatable, where you mis-sketched the distance vs time graph.)

I told you this was right quite a few posts back, before you spun off into wild guesses! http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon6.gif

Now for your acceleration vs time graph ...
 
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  • #39
NascentOxygen said:
You are allowed to say that only once! :smile:

etxjzd.png

Yes, that's about as correct as you are going to manage here. It should sync with the distance vs time graph, so that corresponding times line up. The original distance vs time graph has 3 intervals of roughly equal duration. (That final vertical line is debatable, where you mis-sketched the distance vs time graph.)

I told you this was right quite a few posts back, before you spun off into wild guesses! http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon6.gif

Now for your acceleration vs time graph ...

Ok, well I will trust you. That other guy told me the other graph is correct so I just got really confused. This graph however actually makes sense so I will go with it as it was my instinct.
 
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  • #40
physicsnobrain said:
Ok. But at the end it accelerates slow then quickly accelerates
It might do that, then again it might not. It's difficult to judge from a small sketch. It certainly is not sufficiently obvious that I would agree to it being drawn that way. Neither can I say that the acceleration is fixed and constant here. So the best you can do, without more precise data, is to come up with a rough sketch of what you estimate is happening.
 
  • #41
Ok. So here is the acceleration chart I'm getting off of the velocity chart. Looks good?

2il0t1j.png
 
  • #42
physicsnobrain said:
Ok. So here is the acceleration chart I'm getting off of the velocity chart. Looks good?

2il0t1j.png
I'd give you 4 out of 5 for that. :smile:

During the object's travel, there is a moment which could be described as <CRUNCH!>. Where is this point, and how would it show up on the acceleration vs time plot?

I'd leave out that right hand side vertical line. It's doubtful.
 
  • #43
NascentOxygen said:
I'd give you 4 out of 5 for that. :smile:

During the object's travel, there is a moment which could be described as <CRUNCH!>. Where is this point, and how would it show up on the acceleration vs time plot?

I'd leave out that right hand side vertical line. It's doubtful.

Hmm, I'd say it would show up at both the zero acceleration inbetween and at the end.
 
  • #44
Your answer is not clear.

If you were that object, at what point(s) in time would you be most likely to suffer bruising? Can you put yourself into that scenario?
 
  • #45
NascentOxygen said:
Your answer is not clear.

If you were that object, at what point(s) in time would you be most likely to suffer bruising? Can you put yourself into that scenario?

probably at the very end when the graph ends.
 
  • #46
physicsnobrain said:
probably at the very end when the graph ends.
Explain why.
 
  • #47
NascentOxygen said:
You are allowed to say that only once! :smile:

etxjzd.png

Yes, that's about as correct as you are going to manage here. It should sync with the distance vs time graph, so that corresponding times line up. The original distance vs time graph has 3 intervals of roughly equal duration. (That final vertical line is debatable, where you mis-sketched the distance vs time graph.)

I told you this was right quite a few posts back, before you spun off into wild guesses! http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon6.gif

Now for your acceleration vs time graph ...

Hey guys. Please excuse me. Sorry I didn't notice that the zero slope on the right side of the plateau in the distance time graph was continuous. So this present figure is a better representation of the velocity variation than the one I recommended using in post #12.

Chet
 
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  • #48
NascentOxygen said:
Explain why.

The object instantly goes from having acceleration then goes to 0 acceleration instantly.
 
  • #49
physicsnobrain said:
The object instantly goes from having acceleration then goes to 0 acceleration instantly.
A sudden change in acceleration would not cause discomfort. If you were to step out off a ledge your vertical acceleration would change very quickly from 0 to g, but you would feel no shock. There are two points in the trajectory where something does change suddenly from nonzero to zero, and that would cause a shock. What is that something? If the change were really to take no time at all, what would the acceleration be?
 
  • #50
haruspex said:
A sudden change in acceleration would not cause discomfort. If you were to step out off a ledge your vertical acceleration would change very quickly from 0 to g, but you would feel no shock. There are two points in the trajectory where something does change suddenly from nonzero to zero, and that would cause a shock. What is that something? If the change were really to take no time at all, what would the acceleration be?

Is it when the object moves up, then proceeds to stop moving up instantly?
 
  • #51
NascentOxygen said:
I'd give you 4 out of 5 for that. :smile:

During the object's travel, there is a moment which could be described as <CRUNCH!>. Where is this point, and how would it show up on the acceleration vs time plot?

I'd leave out that right hand side vertical line. It's doubtful.

ok i will try this.
 
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  • #52
So you're saying the acceleration time graph should look like this?
2rfaueb.png
 
  • #53
physicsnobrain said:
The object instantly goes from having acceleration then goes to 0 acceleration instantly.
Except you told me that we don't know what happens there, where I circled at the right. The recording stops as soon as the object passes through zero displacement. We don't know whether it smoothly continues on to a negative displacement or whether it halts at zero (you drew it showing that it stops at zero, but then clarified that you did that in error). So the end of the graph is an unknown; we definitely cannot say there is a bruising crunch there.

What is happening where I drew the red arrow?

EDIT: Ah, I see you answered already:
Is it when the object moves up, then proceeds to stop moving up instantly?
Yes, there is a sudden change of speed there. So how to improve your acceleration vs time graph to show this happening?
 
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  • #54
NascentOxygen said:
EDIT: Ah, I see you answered already:

Yes, there is a sudden change of speed there. So how to improve your acceleration vs time graph to show this happening?

I'm just not seeing it. When I separate this graph into three distinct segments ( reflective of the three in the displacement time graph) this graph seems to make perfect sense to me:
2exrn2b.png


Also, would a good scenario be similar to the one you said previously, but instead of a t-shirt, a wet toilet paper ball is thrown to the ceiling, sticks to the ceiling for a few seconds, then falls back down?
 
  • #55
physicsnobrain said:
I'm just not seeing it. When I separate this graph into three distinct segments ( reflective of the three in the displacement time graph) this graph seems to make perfect sense to me:
2exrn2b.png

Okay. Stop right there. Here's an exam question: an object with some initial upward velocity is subjected to the acceleration you show in your graph. Sketch the expected velocity vs time graph.

Also, would a good scenario be similar to the one you said previously, but instead of a t-shirt, a wet toilet paper ball is thrown to the ceiling, sticks to the ceiling for a few seconds, then falls back down?
Yes. I tried various scenarios, including a sticky tennis ball, wet washing, and a spit ball (bogan kids at school would chew up paper and flick a ball of it onto the classroom ceiling!) That dwell time is important.

Now, your graph for the above?
 

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  • #56
NascentOxygen said:
Okay. Stop right there. Here's an exam question: an object with some initial upward velocity is subjected to the acceleration you show in your graph. Sketch the expected velocity vs time graph.

Now, your graph for the above?

The acceleration time graph I made just seems to work for this, I honestly can't figure out the change:
2dv9c79.png
 
  • #57
During period A, under the influence of a constant acceleration, the velocity is seen to be steadily decreasing. When acceleration disappears by fading to zero, the velocity will remain as it was at that moment. There is nothing to cause it to suddenly drop to zero speed.

So, can you now sketch expected velocity vs time for the acceleration vs time graph you drew?
 
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  • #58
physicsnobrain said:
The acceleration time graph I made just seems to work for this, I honestly can't figure out the change:
2dv9c79.png
If the velocity changes from a nonzero value to zero in an extremely short time, how great is the acceleration?
 
  • #59
haruspex said:
If the velocity changes from a nonzero value to zero in an extremely short time, how great is the acceleration?

not very great. Perhaps if I draw it less on my graph?
 
  • #60
haruspex said:
If the velocity changes from a nonzero value to zero in an extremely short time, how great is the acceleration?


You're saying it should look like this:
2uegugy.png
 
  • #61
No, that is not it.
 
  • #62
Ok, well thanks for the help. The assignment is now due.
 
  • #63
It's this. That crunch cut it short didn't it.

2aijwi1.png
 
  • #64
Also perhaps this could be the scenario. The crunch represents its change from negative acceleration to positive for a brief moment then back to zero.

rvd1mt.png
 
  • #65
If the acceleration only lasted a short time, like you have drawn, then at the end of that short time the object would continue on with a steady speed until some acceleration did again act on it.
 
  • #66
IS IT THIS?? This I think represents it because it can't instantly go to zero velocity right?

suv79f.png
 
  • #67
I think this is accurately representing it, i remember you saying something about the vertical line to the right. Its this isn't it?

suv79f.png
 
  • #68
physicsnobrain said:
Also perhaps this could be the scenario. The crunch represents its change from negative acceleration to positive for a brief moment then back to zero.
You are getting the idea. A short sharp force acted to cause the object to lose all of its speed in a very brief moment. But it's not above the axis, that would cause it to rise up faster.
 
  • #69
NascentOxygen said:
You are getting the idea. A short sharp force acted to cause the object to lose all of its speed in a very brief moment. But it's not above the axis, that would cause it to rise up faster.

So I remove the vertical line like you said before?
suv79f.png
 
  • #70
This is the graph I refer to. You show the spike of acceleration in the right place, but it should not be above the line, it should be below. You need it to act so it slows the object rapidly to stop it. Like bumping your head on an overhead steel beam! <crunch!>

rvd1mt.png
 

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