Simple displacement and acceleration question

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Homework Help Overview

The discussion revolves around a problem involving the calculation of displacement and average velocity for a body over a specified time interval, using the position function s = t^2 - 3t + 2, within the range of 0 to 2 seconds.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the interpretation of the given equation, questioning whether it represents position, velocity, or acceleration. There is confusion regarding the relationship between position and displacement, as well as the need for multiple values of t to determine displacement and average velocity over an interval.

Discussion Status

Participants are actively exploring definitions and clarifying concepts related to position, displacement, and average velocity. Some guidance has been offered regarding the need to consider the entire interval rather than a single value of t, and there is an ongoing examination of the definitions involved.

Contextual Notes

There appears to be a misunderstanding regarding the terms used in the problem, particularly between position and displacement, as well as the distinction between velocity and acceleration. The original poster expresses uncertainty about how to apply the derivative in the context of the problem.

jaydnul
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Homework Statement



Find the body's displacement and average velocity for the given time interval.

s=t^2 -3t +2, 0<=t<=2


Homework Equations



d(displacement)/d(time) is for the acceleration right?


The Attempt at a Solution



So here is my problem. I got 2t-3 for the derivative, but i don't really know how to apply it to the question. I feel like i need just one value for t in order to find the displacement and velocity at that point, right? So how do i incorporate a full interval, 0 to 2? The back of the book says -2 for displacement and -1 for acceleration, but i have no clue how that is the case when you have multiple values of t. I would have just said the 2t-3.
 
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What does the given equation represent? Position? Velocity? Acceleration?

The change in POSITION(not displacement, they are related but not the same) over the change in time is velocity. the change in velocity over the change in time is acceleration.

d(POSITION)/d(time) is not acceleration. When you get confused on these, try and just think about the units you would be working with it. X(meters) over T(seconds) That is m/s which is velocity. I think you are getting position and displacement confused. Position is simply where something is at any given time. Where as the displacement is Δx or the change in its position. Additionally, you are dealing with the CHANGE in time. That being said, yes. It is necessary to have more than one value for t.
 
Last edited:
lundyjb said:
So here is my problem. I got 2t-3 for the derivative, but i don't really know how to apply it to the question. I feel like i need just one value for t in order to find the displacement and velocity at that point, right? So how do i incorporate a full interval, 0 to 2? The back of the book says -2 for displacement and -1 for acceleration, but i have no clue how that is the case when you have multiple values of t. I would have just said the 2t-3.
There's no need for calculus here. You are given the position (s) as a function of time.

What's the definition of displacement?

What's the definition of average velocity? (They want velocity, not acceleration.)
 
What is s(t) when t = 0? What is s(t) when t = 2?
 

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