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Mathematics 20 Online
OpenStudy (wade123):

@Jhannybean

OpenStudy (jhannybean):

Why do you close your questions? :P

OpenStudy (wade123):

for this one i got 16/3

OpenStudy (jhannybean):

\[v(t) = -t^2+4 = 0 \implies t^2 = 4 \implies t = \pm 2 \]So... have any idea how to find the position function? :)

OpenStudy (wade123):

oh wat i set it up wrong, one sec lt me redo it

OpenStudy (jhannybean):

Haha, ok.

OpenStudy (jhannybean):

With what you have given me, you would find the position function by integrating the velocity function.

OpenStudy (jhannybean):

\[s(t) = \int v(t)dt\]

OpenStudy (wade123):

16 (:

OpenStudy (jhannybean):

Im not understanding what "16" means.

OpenStudy (wade123):

16 ft is the distance

OpenStudy (jhannybean):

And can you explain to me what you did to get that?

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