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)..

Question

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)..

anonymous · Answer

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.

anonymous · Answer

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.

he66666 · Answer

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?

he66666 · Answer

plug in*