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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?
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\[3t^{2}\]
not 3t2
yeah if that's the position function
yeah if that's the position function
which it would say r(t) =<\[\sqrt{t}\],\[3t^{2}\]>
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sorry trying to work with equation editor hah
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