Deciphering Velocity-Time Graphs: How to Determine Acceleration?

In summary, the conversation discusses determining acceleration on a velocity-time graph and the confusion over the numbers graphed on the velocity axis. The link provided is used to find the slope of the line and it is suggested to read the page for further clarification. The conversation also touches on whether a changing velocity implies changing acceleration and the method for finding average acceleration.
  • #1
alex2256
6
0
OK, on a velocity-time graph, how does one determine the acceleration?

I actually attempted to find the slope of the line and the graph here:

http://www.physicsclassroom.com/Class/1DKin/U1L4a.cfm (the second graph with the rightward changing velocity) Why does it graph 4, 8, 12, 16 and 20 instead of the 2, 8, 18, 32 and 50 on the velocity axis?

If I find the slope of the line I get 2, 4, 6, 8, 10 which to me looks like constant acceleration and not rightward changing acceleration. Can anybody help me?

Thanks for the help.
 
Physics news on Phys.org
  • #2
alex2256 said:
(the second graph with the rightward changing velocity)

...

looks like constant acceleration and not rightward changing acceleration.

One of these statements does not match the description in the link.
 
  • #3
negitron said:
One of these statements does not match the description in the link.

I'm sorry, I meant "rightward changing velocity". In the second graph on that page from the top, they graph 4, 8, 12, 16 and 20 and not the numbers 2, 8, 18, 32 and 50. Am I missing something blatantly obvious here, or?
 
  • #4
Well, does a changing velocity necessarily imply a changing acceleration?
 
  • #5
Would I need to find the average acceleration by minusing the initial velocity from the final velocity divided by the final time minus the initial time to get 4m/s^2?
 
  • #6
Why? Does the slope of the line change?
 
  • #7
negitron said:
Why? Does the slope of the line change?

Oh.. do you mean I could simply find the slope of the line and that is my acceleration, since the slope of a line is the acceleration of an object on a v-t graph?
 
  • #8
I do suggest you read the linked page in its entirety. You will find your answers there!
 

1. What is a velocity versus time graph?

A velocity versus time graph is a graphical representation of an object's velocity over a period of time. The horizontal axis represents time, while the vertical axis represents velocity.

2. How is velocity calculated from a velocity versus time graph?

Velocity can be calculated by finding the slope of the line on the graph, using the formula velocity = change in position/change in time. The steeper the slope, the greater the velocity.

3. What does a flat line on a velocity versus time graph indicate?

A flat line on a velocity versus time graph indicates that the object is not moving, as the velocity is constant at 0. This could mean that the object is at rest or moving at a constant speed.

4. How can acceleration be determined from a velocity versus time graph?

Acceleration can be determined by finding the slope of the velocity versus time graph. A positive slope indicates positive acceleration, while a negative slope indicates negative acceleration. The steeper the slope, the greater the acceleration.

5. What information can be obtained from a velocity versus time graph?

A velocity versus time graph can provide information about an object's speed, direction of motion, and acceleration. It can also show how an object's velocity changes over time and whether it is moving at a constant speed or accelerating.

Similar threads

  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
505
  • Introductory Physics Homework Help
Replies
2
Views
368
  • Introductory Physics Homework Help
Replies
7
Views
143
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
7K
  • Introductory Physics Homework Help
Replies
4
Views
4K
  • Introductory Physics Homework Help
Replies
3
Views
4K
Back
Top