Find Average Velocity on a Data Graph

Click For Summary

Homework Help Overview

The discussion revolves around finding the average velocity from a distance versus time graph, exploring how to interpret the data presented in such a graph.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the nature of the graph and the method for calculating average velocity, including considerations of slope and weighted averages.

Discussion Status

Some participants have offered guidance on calculating average velocity, while others reflect on their initial assumptions and the complexity of their earlier thoughts. Multiple interpretations of the problem are being explored.

Contextual Notes

There is a mention of potential confusion between distance versus time and velocity versus time graphs, which may affect the understanding of how to calculate average velocity.

physhelp
Messages
7
Reaction score
0
How do you find average velocity by reading off of a data graph?
 
Physics news on Phys.org
That depends, is it a graph of position versus time, velocity versus time, or acceleration versus time?
 
it is d(m) vs t(s)...so it is distance versus time
 
When you have a distance versus time graph the velocity at any given point is the slope of the line at that point. If you have a number of different velocities on the graph you'll want the weighted average, i.e. take each velocity, multiply it by the amount of time the particle had that velocity, add them all up, then divide by the total time.
 
ok thanks a lot
 
It just occurred to me that I made this a lot harder than it needed to be :wink:

I was thinking of the whole matter in terms of a velocity vs. time graph, which is why I was talking about slopes of your distance vs. time graph. The easy way to do this is to simply subtract the initial value for the position from the final value for the position then divide by the time. That's really the same thing as what I said above, except it's computationally much more simple.
 

Similar threads

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