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

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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.
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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?
 
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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?
 
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