Intro Physics homework help - distance and velocity

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Homework Help Overview

The discussion revolves around a physics problem related to distance and velocity, specifically involving calculations from a velocity-time graph. Participants are attempting to clarify their understanding of the problem and the calculations involved.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are sharing their attempts at solving the problem, with one expressing frustration over repeated incorrect answers. Questions arise regarding the correctness of a specific answer and the methods used to arrive at it, including the calculation of areas under the velocity-time graph.

Discussion Status

The discussion is active, with participants engaging in clarifying questions and sharing insights about their calculations. One participant acknowledges a mistake after receiving feedback, indicating a productive exchange of ideas.

Contextual Notes

There appears to be some confusion regarding the interpretation of the velocity-time graph and the calculations involved, which may be affecting the participants' understanding of the problem.

mncyapntsi
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Homework Statement
An elevator going up from one floor to the next, where velocity is positive if it is upward. The acceleration at the start and the deceleration at the end are uniform and of the same magnitude. If the maximum speed of the elevator is 1.26 m/s and it takes a total of 3.40 s for the elevator to move between the floors, what is the distance between the floors if the time that the elevator maintains its top speed is 2.90 s?
Relevant Equations
v=d/t ; x = xo + vt
I have tried this problem over 4 times and keep getting the exact same answer, which is incorrect. Could someone please help by pointing out what I did wrong, or letting me know if I am actually on the completely wrong path?
Thanks!
Have a wonderful day :)
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Is the answer supposed to be ##3.97m##?
 
PeroK said:
Is the answer supposed to be ##3.97m##?
Could I please ask how you got that?
 
mncyapntsi said:
Could I please ask how you got that?
I calculated the area of your velocity-time graph. Whereas, you calculated the area of a velocity-time graph for a speed of ##1.26 m/s## for the full ##3.4s##.
 
OH! I see what I did - thanks a lot!
 

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