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

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The discussion focuses on understanding the difference between distance traveled and displacement in the context of an elevator's velocity graph. A key point is the confusion regarding the distance from t=12-16 seconds, where it is clarified that the distance is -20 m due to the elevator moving downward, despite a positive slope being observed. The average velocity formula used is incorrect for part (d) because the acceleration is not constant, leading to misunderstandings in calculating final position and distance traveled. The user seeks clarification on these concepts, indicating a need for a deeper grasp of the relationship between velocity, displacement, and distance. Overall, the conversation emphasizes the importance of understanding these distinctions in physics problems.
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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