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Mathematics 17 Online
OpenStudy (anonymous):

A weather balloon is released and rises vertically such that its distance s(t) above the ground during the first 10 seconds of flight is given by s(t) = 6 + 2t + t^2, where s(t) is in feet and t is in second. Find the velocity of the balloon at the instant the balloon is 50 ft above the ground.

OpenStudy (anonymous):

v(t)=ds(t)/dt=2+2t s(t)=50 when t^2+2t+6=50 solve for t, and substitute in v(t)

OpenStudy (anonymous):

Yea, I had the first part but I couldn't get the correct answer when finding the solution as I'd get -2 +- sqrt(176) / 2

OpenStudy (anonymous):

Ok, I get x=-1+- 3sqrt(5)

OpenStudy (anonymous):

Lemme try it on a new sheet of paper... Thanks for being so helpful.

OpenStudy (anonymous):

I got it! I think...I'm getting 180 under the square root now...so I think it will come out right.

OpenStudy (anonymous):

\[t=-1+3\sqrt{5}\] is positive from which \[v(t)=2-2+6\sqrt{5} = 6\sqrt{5}\]

OpenStudy (anonymous):

I'm kinda lost now. Where are you getting 2-2+6sqrt5 from? The derivative for s(t) is 2t + 2

OpenStudy (anonymous):

I received the same answer as you for the value of t, though.

OpenStudy (anonymous):

ok ,then 2+2t = 2+2(-1+3sqrt(5))= 2-2+2*3*sqrt(5) = 6sqrt(5)

OpenStudy (anonymous):

where the hell am i messing up? im getting 2sqrt5 instead of 3sqrt5 somehow...that's wrong.

OpenStudy (anonymous):

The equation is x^2+2x-44=0, right?

OpenStudy (anonymous):

Never mind, ignore me...I'm an idiot. I see how you got the answer. Thanks!

OpenStudy (anonymous):

Never call yourself an idiot. Someone might believe you... ;-)

OpenStudy (anonymous):

I'm making things too complex on myself by jumping over steps such as simplifying and stuff...in turn it's causing me to work too hard at getting to the answer and it doesn't APPEAR right even though it is once broken down. Idiot is as idiot does and what I did would be classified as idiotic. :P Good answer given...enjoy your afternoon!

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