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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)..
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0s(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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, 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.

he66666
 5 years ago
Best ResponseYou've already chosen the best response.0So 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?
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