Troubleshooting a Velocity vs. Time Graph: Position at t=10

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The discussion revolves around calculating the position from a velocity vs. time graph, starting at 15 m/s and transitioning to -10 m/s at t=4, then returning to 15 m/s at t=10. The initial calculations involved finding the slopes of the segments and integrating to determine the area under the curve, resulting in a total position of 25m. However, the user encountered an error message indicating the answer was incorrect, leading to questions about significant figures and potential errors in the problem setup. Despite confirming the calculations were correct, the possibility of an error in the electronic system was suggested. The conversation emphasizes the importance of careful calculations and the potential for discrepancies in online problem sets.
Psyguy22
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We are given a velocity vs. time graph. It starts at 15 m/s at t=0 and goes to -10m/s at t=4
then it goes from -10m/s at t=4 to 15m/s at t=10 and we are told to find the postion after t=10

I first found the slope of the first segment which was -6.25. so I did ∫-6.25t+15 dt from 0 to 4 and got 10m
I found the second slope, 4.16..., and did ∫4.16...t-10 dt from 0-6 (because I moved it to the orgin so it made sense) and got 15m. So I added the two segments to get 25m. But the site keeps saying that's the wrong answer. What am I doing wrong?
 
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Does the sight care about significant figures? Hard to believe ∫4.16t-10 dt comes out to exactly 15m.
 
It comes to 15 when you put 25/6 which is 4.16 repeating. But yes the site does care about significant figures.
 
oh, I didn't do the math myself I just took it as 4.16 flat. Well I did the math myself as well as taking the area under the curve of the graph and got the same answer as you both ways. Honestly at this point I would say they possibly put in a wrong answer for the problem since it's electronic. I know that's happened to people I know before so you might as well ask.
 
ok thank you for your help!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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