The discussion revolves around the interpretation of the area under a velocity-time (v-t) graph, particularly when the integral results in zero. Participants explore why a zero integral does not necessarily indicate that a particle has returned to its original position, focusing on concepts of displacement, motion phases, and the implications of velocity being positive or negative.
Participants express differing views on the implications of a zero integral in a v-t graph. There is no consensus on whether this indicates a return to the origin or simply a return to the starting point, leading to an unresolved discussion.
The discussion highlights limitations in understanding displacement versus total distance traveled, as well as the effects of changing velocity on the final position. There are also unresolved assumptions regarding the dimensionality of motion.
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?
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
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?