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Mathematics 12 Online
OpenStudy (anonymous):

to find the acceleration vector of x=t√ and y=3t2−2t when t=1 would you just take the derivative of both x and y twice?

OpenStudy (anonymous):

\[3t^{2}\]

OpenStudy (anonymous):

not 3t2

OpenStudy (anonymous):

yeah if that's the position function

OpenStudy (anonymous):

yeah if that's the position function

OpenStudy (anonymous):

which it would say r(t) =<\[\sqrt{t}\],\[3t^{2}\]>

OpenStudy (anonymous):

sorry trying to work with equation editor hah

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