Kinematics 4 -- Calculate the velocity from the position versus time graph

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SUMMARY

This discussion focuses on calculating average velocity from a position versus time graph in kinematics. The key equation used is vav = total distance traveled / total time taken. Participants suggest using numerical methods such as Newton's forward interpolation or Lagrange's interpolation to analyze displacement-time relationships over specified intervals. The average velocity can be interpreted as the slope of a line on the graph, with a specific example indicating that the instantaneous velocity at t = 10 s corresponds to a time moment of 16 s.

PREREQUISITES
  • Understanding of kinematic equations, specifically average velocity calculation.
  • Familiarity with numerical methods such as Newton's forward interpolation.
  • Knowledge of Lagrange's interpolation for analyzing displacement-time relationships.
  • Ability to interpret graphical data, particularly slopes of lines on position versus time graphs.
NEXT STEPS
  • Research "Newton's forward interpolation method" for numerical analysis in kinematics.
  • Study "Lagrange's interpolation" for advanced techniques in data fitting.
  • Learn how to derive average velocity from graphical data in physics.
  • Explore the concept of instantaneous velocity and its graphical representation.
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Students studying physics, particularly those focusing on kinematics, as well as educators and tutors looking to enhance their teaching methods in velocity calculations.

Pushoam
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Homework Statement


upload_2017-7-11_21-51-59.png


Homework Equations


vav = total distance traveled/ total time taken

The Attempt at a Solution


I have a problem in part c .
How am I supposed to solve it?
Is there any standard way to solve it?
 

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Take an arbitrary time to. How can you graphically determine the average velocity for the time interval between t = 0 and t = to? [Edit: Can you interpret the average velocity as the slope of some line?]

It might help to first take a specific example. Suppose you take t0 = 10 s. How would you get the average velocity for the time interval t = 0 to t = 10 s? Does this average velocity equal the instantaneous velocity at t = 10 s?
 
Last edited:
Pushoam said:
I have a problem in part c .
How am I supposed to solve it?
Is there any standard way to solve it?
Irodov?

Pushoam said:
Is there any standard way to solve it?
One standard way could be using numerical methods. Newton's forward interpolation or Lagrange's interpolation will work.
You can find the displacement-time relationship for three intervals: i)from t=0 to t=10s ii)from t=10 to t=14.5s iii) from t=14.5 to t=20s.
Once you have the s-t relationship, you can find t0.
 
As TSny indicates, you can pretty much determine the answer for part c by eye, just by looking at the graph.
 
TSny said:
Edit: Can you interpret the average velocity as the slope of some line?]
Thanks.
I got it.
I never interpreted average velocity as the slope of some line. Thanks for this insight.
So, what I have to find out is the time moment at which the line whose slope defines the instantaneous velocity passes through the origin. This time moment turns out to be 16 s.
cnh1995 said:
Irodov?

Yes

Thanks to all for replying.
 
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Pushoam said:
So, what I have to find out is the time moment at which the line whose slope defines the instantaneous velocity passes through the origin. This time moment turns out to be 16 s.
Yes. Good work.
 
Thanks, TSny
 

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