Displacement as Function of Time Graph Question

Click For Summary
SUMMARY

The forum discussion centers on constructing velocity and acceleration graphs from a given displacement vs. time graph. Users discussed the proper methods for deriving these graphs, emphasizing the importance of accurately interpreting the slopes of the displacement graph. A scenario involving a helicopter rescuing a kitten was proposed to illustrate the motion described by the graphs. Participants provided feedback on the accuracy of the graphs, highlighting the need for precise labeling and the correct representation of changes in velocity and acceleration.

PREREQUISITES
  • Understanding of displacement, velocity, and acceleration concepts
  • Ability to interpret and construct graphs based on mathematical relationships
  • Familiarity with tangential lines and their significance in graph analysis
  • Knowledge of basic physics principles related to motion
NEXT STEPS
  • Learn how to derive velocity from displacement graphs using tangential lines
  • Study the relationship between acceleration and the slope of velocity graphs
  • Explore real-life applications of motion graphs in physics
  • Practice constructing and analyzing motion graphs with varying complexities
USEFUL FOR

Students studying physics, educators teaching motion concepts, and anyone interested in understanding the graphical representation of motion dynamics.

  • #61
No, that is not it.
 
Physics news on Phys.org
  • #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
 
  • #71
NascentOxygen said:
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


Ahhh, so its slow acting. SO like this:

rizrph.png
 
  • #72
physicsnobrain said:
Ahhh, so its slow acting. SO like this:
When you bump your head on an overhead beam, it doesn't feel like a gentle slow-acting stop to me!

Your graph is improving. Not far to go now ...
 
  • #73
Ahhh, it must be this then.

2qa4r5s.png
 
  • #74
physicsnobrain said:
Ahhh, it must be this then.

2qa4r5s.png
Not quite, but you are close. It's not as complicated as you show. But you have had enough help with this assignment that I think that should do for now. If you think about it more carefully, you might be able to get 100%. Post again if you come up with a better plot.

Remember that I said that vertical line on the extreme right is questionable, for the acceleration vs time plot, as I explained earlier. I would omit it.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
497
  • · Replies 6 ·
Replies
6
Views
1K
Replies
6
Views
2K
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 14 ·
Replies
14
Views
5K
  • · Replies 1 ·
Replies
1
Views
9K
  • · Replies 11 ·
Replies
11
Views
3K