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anonymous
 one year ago
Help!
anonymous
 one year ago
Help!

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1is your function like this: \[\Large s\left( t \right) = \sqrt {2t + 1} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1In order to find the acceleration, we have to compute the second derivative of that function s(t)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I can rewrite your function as below: \[\Large s\left( t \right) = \sqrt {2t + 1} = {\left( {2t + 1} \right)^{1/2}}\] so the first derivative is: \[\Large \frac{{ds}}{{dt}} = \frac{1}{2} \cdot {\left( {2t + 1} \right)^{\frac{1}{2}  1}} \cdot 2 = ...?\] please complete

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you guys know what website I can take screen shots with?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Michele can you help me with a question???

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0When I derive it I got 1/sqr root(2t+1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Please don't ask questions on my question :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And I would plug in 4 into this?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large \frac{{ds}}{{dt}} = \frac{1}{2} \cdot {\left( {2t + 1} \right)^{\frac{1}{2}  1}} \cdot 2 = \frac{1}{{\sqrt {2t + 1} }}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry but Michele can you help me with a question I will fan and medal

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now, we have to compute the second derivative of your starting function, namely we have to compute this: \[\Large \frac{{{d^2}s}}{{d{t^2}}} = \frac{d}{{dt}}\left( {\frac{1}{{\sqrt {2t + 1} }}} \right)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay I got 1/(2t+1)^(3/2)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we can rewrite that last expression as below: \[\Large \begin{gathered} \frac{{{d^2}s}}{{d{t^2}}} = \frac{d}{{dt}}\left( {\frac{1}{{\sqrt {2t + 1} }}} \right) = \hfill \\ \hfill \\ = \frac{d}{{dt}}{\left( {2t + 1} \right)^{  1/2}} =  \frac{1}{2} \cdot {\left( {2t + 1} \right)^{  \frac{1}{2}  1}} \cdot 2 = ...? \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1correct! Now, you can set t=4, into that last second derivative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much for your help!!
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