How Do You Convert a Velocity-Time Graph to a Position-Time Graph?

  • Thread starter Thread starter 1irishman
  • Start date Start date
  • Tags Tags
    Graph
Click For Summary
SUMMARY

The discussion focuses on converting a velocity-time (v-t) graph to a position-time (p-t) graph by analyzing the area under the v-t graph. The area represents the displacement, which is crucial for calculating the greatest distance from the origin. Participants emphasize understanding the relationship between velocity, time, and distance to effectively interpret the graph. Key concepts include the significance of the area under the curve and how it translates to position over time.

PREREQUISITES
  • Understanding of basic calculus concepts, specifically integration.
  • Familiarity with kinematic equations in physics.
  • Knowledge of graph interpretation, particularly velocity-time graphs.
  • Basic understanding of displacement and distance in physics.
NEXT STEPS
  • Study the principles of integration to calculate areas under curves.
  • Learn how to derive position-time equations from velocity-time graphs.
  • Explore kinematic equations and their applications in motion analysis.
  • Practice converting various v-t graphs to p-t graphs using real-world examples.
USEFUL FOR

Students studying physics, educators teaching motion concepts, and anyone interested in graph analysis related to kinematics.

1irishman
Messages
243
Reaction score
0

Homework Statement


Can anyone tell me how to calculate the greatest distance from the origin given a v-t graph?



Homework Equations


i don't know.


The Attempt at a Solution


no idea.
 
Physics news on Phys.org
1irishman said:

Homework Statement


Can anyone tell me how to calculate the greatest distance from the origin given a v-t graph?



Homework Equations


i don't know.


The Attempt at a Solution


no idea.
What is distance in terms of v and t? What does the area under the v-t graph represent?

AM
 
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?

Similar threads

  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 13 ·
Replies
13
Views
3K
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
9K