Understanding a Velocity-Time Graph

In summary,I think we are still in the earlier parts of Physics and I am confused at how "values" work for a velocity-time graph. We are using the formulas to solve an area of a triangle and rectangle to find the total displacement. If a diagonal line begins from above and continue to go down, that denotes a negative velocity but in solving for the height of a triangle, will it also be negative?
  • #1
PleaseAnswerOnegai
13
1
Summary:: I think we are still in the earlier parts of Physics and I am confused at how "values" work for a velocity-time graph. We are using the formulas to solve an area of a triangle and rectangle to find the total displacement. If a diagonal line begins from above and continue to go down, that denotes a negative velocity but in solving for the height of a triangle, will it also be negative?

4.JPG

For the area of triangle, we use " 1/2(b)(h) ". For the first triangle will its height be a -4 in solving, as the line is going down therefore it is decelerating, or should I use a positive 4? How about the two other triangles in the trapezoid? How about the rectangle, will its height be -4?
We were asked to solve the displacement within the time frame of (0,9) so the triangle beyond that is not a concern for me at the moment.
 
Physics news on Phys.org
  • #2
What is displacement?

What is the relationship between velocity and displacement?

Can you sketch a displacement vs time graph from the velocity vs time graph?
 
  • #3
PeroK said:
What is displacement?

What is the relationship between velocity and displacement?

Can you sketch a displacement vs time graph from the velocity vs time graph?
Based on my understanding:

Displacement is a scalar measurement that quantifies how far we have gone from an original point/place/position relative to direction as opposed to Distance which simply measures, well, total "distance."

Similarly, velocity takes into account direction. And its relationship with displacement tells us how fast you have gone in any specified direction relative to where you began.

That is why we can have negative velocity and displacement as that would mean you went the "opposite direction"; backward instead of forward.

I think yes, once you've found the displacement.

I may be wrong but please correct me, thank you very much.
 
  • #4
PleaseAnswerOnegai said:
Based on my understanding:

Displacement is a scalar measurement that quantifies how far we have gone from an original point/place/position relative to direction as opposed to Distance which simply measures, well, total "distance."

Similarly, velocity takes into account direction. And its relationship with displacement tells us how fast you have gone in any specified direction relative to where you began.

That is why we can have negative velocity and displacement as that would mean you went the "opposite direction"; backward instead of forward.

I think yes, once you've found the displacement.

I may be wrong but please correct me, thank you very much.
Displacement is a vector quantity. Distance is a scalar.

Velocity is the rate of change of displacement.

For your example, calculate:

The displacement at ##1s, 2s, 3s##.
 
  • #5
Pardon me, I mixed them up.
 
  • #6
In solving of the triangle, should my solution be 1/2(1)(-4)?
 
  • #7
PleaseAnswerOnegai said:
In solving of the triangle, should my solution be 1/2(1)(-4)?
PeroK said:
For your example, calculate:

The displacement at ##1s, 2s, 3s##.

Why don't you calculate the displacement at ##1s, 2s, 3s##?

If in doubt calculate!
 
  • #8
PeroK said:
Why don't you calculate the displacement at ##1s, 2s, 3s##?

If in doubt calculate!
Would the displacement at 1s be 2 cm?
 
  • #9
PleaseAnswerOnegai said:
Would the displacement at 1s be 2 cm?
Why ##cm##?
 
  • #10
PeroK said:
Why ##cm##?
I'm really sorry, I meant 2m. Would that be correct?
 
  • #11
PleaseAnswerOnegai said:
I'm really sorry, I meant 2m. Would that be correct?
Yes.
 
  • #12
And at 2s, the displacement would be 0m because of 2m + (-2m)?
 
  • Like
Likes PeroK
  • #13
PleaseAnswerOnegai said:
And at 2s, the displacement would be 0m because of 2m + (-2m)?
Yes. In summary:

When the velocity is positive, the graph is above the ##t## axis and area represents a positive displacement. Regardless of the sign of the acceleration.

When the velocity is negative, the graph is below the ##t## axis and area represents a negative displacement.

Note: if you have studied calculus, the integral takes care of this for you. If you integrate a velocity vs time function, then the integral automatically calculates anything below the ##t## axis as negative.

If you calculate the areas by hand, you have to put in the negative signs for anything below the ##t## axis.

Note that that signs on your original diagram were the wrong way round. As you now know.
 
  • #14
PeroK said:
Yes. In summary:

When the velocity is positive, the graph is above the ##t## axis and area represents a positive displacement. Regardless of the sign of the acceleration.

When the velocity is negative, the graph is below the ##t## axis and area represents a negative displacement.

Note: if you have studied calculus, the integral takes care of this for you. If you integrate a velocity vs time function, then the integral automatically calculates anything below the ##t## axis as negative.

If you calculate the areas by hand, you have to put in the negative signs for anything below the ##t## axis.

Note that that signs on your original diagram were the wrong way round. As you now know.
Thank you very much! This helps me greatly. And I have taken Calculus, that would make sense. Thank you again and have a nice day :)

Oh, and I wouldn't want to get complacent hahaha, upon calculating: the total displacement in 9s would be -24m, correct?
 
  • #15
PleaseAnswerOnegai said:
Oh, and I wouldn't want to get complacent hahaha, upon calculating: the total displacement in 9s would be -24m, correct?
Correct.

If you make separate triangles for times where the velocity is positive and regions where it is negative then it's easier to keep track of the signs.

In some cases symmetry arguments work as well. Between t=0 and t=2 for example: As you see both triangles cancel each other to get 0 displacement. Knowing the area of the triangles isn't important as long as you know the areas are the same (but on opposite sides of the x axis).
 

Related to Understanding a Velocity-Time Graph

What is a velocity-time graph?

A velocity-time graph is a graphical representation of an object's velocity over a period of time. It shows how an object's velocity changes over time, with velocity on the y-axis and time on the x-axis.

How do you interpret a velocity-time graph?

The slope of a velocity-time graph represents the object's acceleration. A positive slope indicates a positive acceleration (speeding up), while a negative slope indicates a negative acceleration (slowing down). The area under the graph represents the object's displacement.

What does a flat line on a velocity-time graph indicate?

A flat line on a velocity-time graph indicates that the object is not moving, as there is no change in velocity over time. This could mean that the object is at rest or moving at a constant velocity.

How can you calculate the acceleration from a velocity-time graph?

The acceleration can be calculated by finding the slope of the velocity-time graph. This can be done by selecting two points on the graph and using the formula: acceleration = change in velocity / change in time.

What are some real-life applications of velocity-time graphs?

Velocity-time graphs are commonly used in physics to analyze the motion of objects, such as cars, trains, and projectiles. They are also used in sports to track the performance of athletes, and in engineering to design and test vehicles and machines.

Similar threads

Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
765
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
135
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
956
Back
Top