Integral of a position v time graph; meaningless?

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SUMMARY

The discussion centers on the integration of a position vs. time graph, specifically questioning the meaning of the resulting units, which are meters seconds. Participants clarify that integrating a velocity vs. time graph yields position, while integrating a position vs. time graph does not produce a standard physical quantity. The conversation highlights the confusion surrounding units like watt hours and their relationship to joules, emphasizing that watt hours represent energy, not position. The integration process is linked to calculating average position over time.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques.
  • Familiarity with kinematic concepts, particularly velocity and position.
  • Knowledge of physical units, including joules and watt hours.
  • Basic grasp of energy concepts in physics.
NEXT STEPS
  • Explore the concept of integration in calculus, focusing on physical applications.
  • Research the relationship between energy units, specifically joules and watt hours.
  • Learn about kinematic equations and their applications in physics.
  • Investigate average value theorems in calculus and their implications in physics.
USEFUL FOR

Students of physics, educators teaching calculus and kinematics, and anyone interested in understanding the implications of integrating different types of graphs in physical contexts.

pakmingki
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So, you have a graph of velocity vs. time. You integrate, you get position. What if you have a position vs time graph? For example, meters vs. seconds. If you integrate it, you will get meters seconds. Now, does this mean anything? For example, I've heard of "watt hours" but i simply don't understand how this makes any sense. What would it mean to integrate a position vs time graph?
 
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One reason you might want to integrate it is to find the average position.

<x>=1/T Int[x(t)dt]
 
Watt hours (more commonly seen as kilowatt hours--kwh) are a unit of energy.

Try multiplying the units for watts by the units for hours and see what you arrive with.

(J/t) * t = Joules
 
o
why not just call them joules then?
ive seen products, such as surge protectors labeled with, "handles up to 16,000 joules."
 
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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