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is your function like this: \[\Large s\left( t \right) = \sqrt {2t + 1} \]
Yes
In order to find the acceleration, we have to compute the second derivative of that function s(t)

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I 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
do you guys know what website I can take screen shots with?
Michele can you help me with a question???
When I derive it I got 1/sqr root(2t+1)
correct!
Please don't ask questions on my question :)
And I would plug in 4 into this?
\[\Large \frac{{ds}}{{dt}} = \frac{1}{2} \cdot {\left( {2t + 1} \right)^{\frac{1}{2} - 1}} \cdot 2 = \frac{1}{{\sqrt {2t + 1} }}\]
Sorry but Michele can you help me with a question I will fan and medal
now, we have to compute the second derivative of your starting function, namely we have to compute this: \[\Large \frac{{{d^2}s}}{{d{t^2}}} = \frac{d}{{dt}}\left( {\frac{1}{{\sqrt {2t + 1} }}} \right)\]
Okay I got -1/(2t+1)^(3/2)
we can rewrite that last expression as below: \[\Large \begin{gathered} \frac{{{d^2}s}}{{d{t^2}}} = \frac{d}{{dt}}\left( {\frac{1}{{\sqrt {2t + 1} }}} \right) = \hfill \\ \hfill \\ = \frac{d}{{dt}}{\left( {2t + 1} \right)^{ - 1/2}} = - \frac{1}{2} \cdot {\left( {2t + 1} \right)^{ - \frac{1}{2} - 1}} \cdot 2 = ...? \hfill \\ \end{gathered} \]
correct! Now, you can set t=4, into that last second derivative
Okay I got -1/27
that's right!
Thank you so much for your help!!
:)

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