How do I calculate the instantaneous velocity

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SUMMARY

The discussion focuses on calculating instantaneous velocity (vinst) from a distance-time (d-t) graph. Participants confirm that vinst is determined by taking the first derivative of the movement equation at specific points on the graph. Slope breaks, where the graph changes direction, are identified as critical points for calculating vinst. The use of tangent lines to find the slope at these points is emphasized, although initial attempts at rise over run calculations yielded incorrect results.

PREREQUISITES
  • Understanding of calculus, specifically derivatives
  • Familiarity with distance-time graphs
  • Knowledge of how to identify slope breaks on a graph
  • Ability to compute rise over run for tangent lines
NEXT STEPS
  • Study the concept of derivatives in calculus
  • Learn how to graph movement equations accurately
  • Explore methods for identifying slope breaks in various functions
  • Practice calculating instantaneous velocity using different movement equations
USEFUL FOR

Students studying calculus, physics enthusiasts, and anyone interested in understanding motion analysis through graphical representation.

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I've constructed a d-t graph and the question is, how do I calculate the instantaneous velocity, vinst, for each slope break? How do I find the slope break? Do I calculate it using v=chginx/chgint formed by the tangent line to the point or using the point alone simular to Vavg. I've drawn tangent lines to the peaks of the curves and computed rise over run with incorrect results.
 
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Have you taken calculus, or do you know how to take derivatives?
 
vinst

you have the movement equation? if you have it and you have taken calculus you must know the vinst ist the first derivative of the function evaluated at the point you want to calculate vinst.
 
Thanks. I was puzzled as to where to find the slope breaks. Once I graphed this out, I saw several linear lines that crossed each other along the graph and it was at these points I calculated Vinst.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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