anonymous
  • anonymous
Help!
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
WOULDN'T YOU JUST PLUG THEM IN? \[3(3)^{2}-4(3)+1\]
anonymous
  • anonymous
No because thats not an option
Michele_Laino
  • Michele_Laino
you have to compute the first derivative of the function v(t) at t=3, namely: \[\Large a\left( t \right) = \frac{d}{{dt}}\left( {3{t^2} - 4t + 1} \right) = ...?\]

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anonymous
  • anonymous
6t-4
Michele_Laino
  • Michele_Laino
ok! now replace t with 3, what do you get?
anonymous
  • anonymous
14
Michele_Laino
  • Michele_Laino
correct! your answer is a=14 feet/sec^2
anonymous
  • anonymous
Thank you so much!
anonymous
  • anonymous
michele is right cause accelration is the rate of change of velocity per time!
Michele_Laino
  • Michele_Laino
thanks!! :)
Michele_Laino
  • Michele_Laino
thanks!! @Hassan3130
anonymous
  • anonymous
Thank you @Hassan3130 !

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