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he66666

  • 5 years ago

The position function of an object that moves in a straight line is s(t) = 1 + 2t - [8/(t^2 +1)], 0≤t≤2. Calculate the maximum and minimum velocities of the object over the given time interval. Answer: min is 2, max is 2 + 3√3. How do you solve this? I know that you have to find the second derivative. When I do that though, I get t=±√(1/3)..

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  1. anonymous
    • 5 years ago
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    s(t) = \[1+2t-(8/t^2+1)\] I am working through the problem to ensure I am doing it correctly. I will be with you in a moment.

  2. anonymous
    • 5 years ago
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    Sorry, I was interrupted. From what I can tell, there is only a maximum velocity from interval t=0 to t=2, and you were correct in saying it was sqrt1/3.

  3. he66666
    • 5 years ago
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    So I would just have to plu gin sqrt1/3 into the original equation and that would be my maximum? Does this mean that the answer is wrong?

  4. he66666
    • 5 years ago
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    plug in*

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