Intro Physics homework help - distance and velocity

AI Thread Summary
The discussion revolves around a physics homework problem related to distance and velocity, where the original poster repeatedly arrives at an incorrect answer of 3.97m. They seek assistance in identifying their mistake or confirming their approach. Another participant clarifies that they calculated the area of the velocity-time graph differently, leading to the confusion. The original poster realizes their error in calculating the area for a speed of 1.26 m/s over the full time of 3.4 seconds. This exchange highlights the importance of accurately interpreting velocity-time graphs in physics problems.
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 :)
Screen Shot 2021-09-14 at 2.59.30 PM 1.png
 
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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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