Why Does Zero Integration of a Velocity-Time Graph Not Confirm Return to Origin?

  • Thread starter Thread starter The_Engineer
  • Start date Start date
  • Tags Tags
    Kinematics
AI Thread Summary
A zero integration of a velocity-time graph indicates that the total displacement is zero, but it does not confirm that the particle returned to the origin. The particle could have moved in one direction and then reversed, resulting in a net displacement of zero without necessarily being at the origin. To determine the final position, one must consider the areas under the velocity curve, which represent displacement in both positive and negative directions. Integrating the absolute value of the velocity provides total distance traveled, while integrating the velocity itself gives the net displacement. Thus, the final position depends on the initial position and the net displacement calculated from the velocity-time graph.
The_Engineer
Messages
17
Reaction score
0
Say you have a v-t diagram for the motion of a particle in one dimension where the velocity is positive at first and then negative later. If you integrate and get zero, why doesn't that mean that the particle started moving and then came back to the origin?
 
Last edited:
Physics news on Phys.org
The_Engineer said:
If a particle undergoes multiple phases of motion (ex. accelerating, then decelerating, then constant acceleration, etc..) then how can you determine where the particle's final position is?

I'm imagining a v-t diagram and if you get the area under all of the velocity curves then you get the total displacement, but not the final position (the particle may have been moving backwards...) How do you get the final position?

Let's keep it simple and apply this only to one dimension.


EDIT: Does integrating the absolute value of all the velocity equations yield total displacement while just integrating yields the final position?

Are we to assume that all motion is in one dimension?

If the particle is moving backwards the velocity will be negative so the change in displacement during that period, ∫vdt, will be negative.

Displacement is the distance from the origin with its direction from the origin (ie. + or - x). The change in displacement is defined as the final displacement (position) minus the initial displacement .

AM
 
Andrew Mason said:
Are we to assume that all motion is in one dimension?

If the particle is moving backwards the velocity will be negative so the change in displacement during that period, ∫vdt, will be negative.

Displacement is the distance from the origin with its direction from the origin (ie. + or - x). The change in displacement is defined as the final displacement (position) minus the initial displacement .

AM

Yes, assuming that the motion is in one dimension, does an integral of zero of a v-t curve indicate that the particle has traveled back to the origin or hasn't moved at all?
 
It has traveled back to its starting point, which may or may not be the origin.
 
The_Engineer said:
Say you have a v-t diagram for the motion of a particle in one dimension where the velocity is positive at first and then negative later. If you integrate and get zero, why doesn't that mean that the particle started moving and then came back to the origin?

Has anyone actually suggested it doesn't?
 

Thread '1:1 versus 2:1 Mechanical Advantage debate'

The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
View full post »

Thread 'Newton's first law?'

Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
View full post »
Thread 'Beam on an inclined plane'

Thread 'Beam on an inclined plane'

Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
View full post »

Similar threads

Why is the integrated velocity times change in velocity zero?
Replies
3
Views
1K
Free body diagrams, coordinate systems origin/orientation
Replies
14
Views
3K
Mechanics ( velocity - time graph )
Replies
15
Views
4K
What does time=0 represent in kinematics
Replies
3
Views
2K
Position vs. time graph and the derivative
Replies
5
Views
3K
Velocity of a particle at time t in a rotating frame
Replies
17
Views
2K
Quick Question about Pendulum Graphs
Replies
8
Views
7K
Why would this be accelerating positively?
Replies
6
Views
910
Does Reversing Time Affect Acceleration and Velocity in Classical Physics?
Replies
3
Views
1K
Finding the location of a kinematic particle in motion based on time.
Replies
1
Views
2K

Hot Threads

Recent Insights

Back
Top