Graph of Elevator's Velocity: Understanding Displacement and Distance Traveled

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Homework Help Overview

The discussion revolves around understanding the concepts of distance traveled and displacement in the context of an elevator's velocity graph. Participants are exploring the differences between these two concepts as they relate to specific time intervals and calculations involving initial and final positions.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are questioning the interpretation of distance versus displacement, particularly regarding the sign of the distance traveled during specific time intervals. There is also an exploration of the equations used to calculate distance and final position, with some participants expressing confusion about their setups and the implications of non-constant acceleration.

Discussion Status

The discussion is active, with participants clarifying their understanding of the problem and questioning their approaches. Some guidance has been offered regarding the nature of average velocity and the implications of acceleration not being constant. However, there is no explicit consensus on the correct interpretation of distance versus displacement.

Contextual Notes

Participants are navigating the constraints of the problem, including the specific requirements of parts (c) and (d) of the homework, which ask for distance traveled and final position, respectively. There is an acknowledgment of potential misunderstandings in the application of equations due to the changing acceleration of the elevator.

reminiscent
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Homework Statement


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Homework Equations

The Attempt at a Solution


This is more of a help with understanding problem. I understand a) and b), but for c), I don't understand:
1. Why is the distance from t=12-16 -20 m and not +20 m? Isn't there a positive slope? I thought this question was asking for distance traveled, not displacement?
2. My attempt at c) was to use the equation y(t)-y0 = 1/2(v0+v(t))t, but my answer was 325 m. y0=+75 m, v0=20 m/s, v(t)=5 m/s, and t=20s. What did I do wrong?
 
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Part (c) asks for distance traveled, but part (d) asks for the final position.
 
TSny said:
Part (c) asks for distance traveled, but part (d) asks for the final position.
So what did I do wrong in the equation I set up in d)?
 
reminiscent said:
y(t)-y0 = 1/2(v0+v(t))t

The average velocity is not 1/2(v0+v(t)) for part (d) because the acceleration is not constant over the entire time interval.
 
TSny said:
The average velocity is not 1/2(v0+v(t)) for part (d) because the acceleration is not constant over the entire time interval.
Oh okay, I understand now. Thanks. But what about the distance from t=12-16 seconds? Why is it -20 m and not +20 m?
 

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