Finding Instantaneous Acceleration in a Velocity-Time Graph

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SUMMARY

The discussion centers on determining instantaneous acceleration from a velocity-time graph. Participants clarify that instantaneous velocity at points A and B can be directly read from the graph, as it represents velocity (v) versus time (t). However, instantaneous acceleration is not well-defined at these points due to abrupt changes in acceleration. A suggestion is made to create a scatter plot of velocity points and find the line of best fit to calculate average velocity, which may aid in understanding acceleration over time intervals.

PREREQUISITES
  • Understanding of velocity-time graphs
  • Knowledge of instantaneous velocity and acceleration concepts
  • Familiarity with slope calculations in graphs
  • Basic skills in creating scatter plots and lines of best fit
NEXT STEPS
  • Learn how to calculate instantaneous acceleration from a velocity-time graph
  • Study the concept of tangent lines and their significance in physics
  • Explore techniques for creating scatter plots and determining lines of best fit
  • Investigate the implications of discontinuities in acceleration on motion analysis
USEFUL FOR

Students and educators in physics, particularly those focusing on kinematics, as well as anyone interested in graph analysis and motion dynamics.

welai
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http://i52.tinypic.com/95zrsw.png

^ okay, if my velocity-time graph looks like that (it was a quick sketch), and I need to find the INSTANTANEOUS velocity at point A and point B, how do I do it?

I mean, I understand the slope of the tangent = instantaneous acceleration, but this is not a curve. Thus, I also understand to use the normal straight slope. But I don't understand, WHICH slope is the INSTANTANEOUS acceleration for those points?
 
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welai said:
http://i52.tinypic.com/95zrsw.png

^ okay, if my velocity-time graph looks like that (it was a quick sketch), and I need to find the INSTANTANEOUS velocity at point A and point B, how do I do it?

I mean, I understand the slope of the tangent = instantaneous acceleration, but this is not a curve. Thus, I also understand to use the normal straight slope. But I don't understand, WHICH slope is the INSTANTANEOUS acceleration for those points?

You read the instantaneous velocities at A and B directly from the graph (because your plot is v vs. t). On the other hand, if you meant to say instantaneous "acceleration" (not velocity), then at A and B there is *no well-defined value*: the acceleration changes instantly from one constant value to another, so the acceleration at one 100 billionth of a second before A is different than the acceleration at one 100 billionth of a second after A.

RGV
 


Ray Vickson said:
You read the instantaneous velocities at A and B directly from the graph (because your plot is v vs. t). On the other hand, if you meant to say instantaneous "acceleration" (not velocity), then at A and B there is *no well-defined value*: the acceleration changes instantly from one constant value to another, so the acceleration at one 100 billionth of a second before A is different than the acceleration at one 100 billionth of a second after A.

RGV

Thank you! Just another simple question, then would it be easier if I make a scatter-plot of my velocity points on my vt graph, then I'll find the line of best fit for average velocity, then find the instantaneous acceleration at the time interval of A and B?
 

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