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## anonymous one year ago Help!

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1. Michele_Laino

is your function like this: $\Large s\left( t \right) = \sqrt {2t + 1}$

2. anonymous

Yes

3. Michele_Laino

In order to find the acceleration, we have to compute the second derivative of that function s(t)

4. Michele_Laino

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

5. anonymous

do you guys know what website I can take screen shots with?

6. anonymous

Michele can you help me with a question???

7. anonymous

When I derive it I got 1/sqr root(2t+1)

8. Michele_Laino

correct!

9. anonymous

Please don't ask questions on my question :)

10. anonymous

And I would plug in 4 into this?

11. Michele_Laino

$\Large \frac{{ds}}{{dt}} = \frac{1}{2} \cdot {\left( {2t + 1} \right)^{\frac{1}{2} - 1}} \cdot 2 = \frac{1}{{\sqrt {2t + 1} }}$

12. anonymous

Sorry but Michele can you help me with a question I will fan and medal

13. Michele_Laino

now, we have to compute the second derivative of your starting function, namely we have to compute this: $\Large \frac{{{d^2}s}}{{d{t^2}}} = \frac{d}{{dt}}\left( {\frac{1}{{\sqrt {2t + 1} }}} \right)$

14. anonymous

Okay I got -1/(2t+1)^(3/2)

15. Michele_Laino

we can rewrite that last expression as below: $\Large \begin{gathered} \frac{{{d^2}s}}{{d{t^2}}} = \frac{d}{{dt}}\left( {\frac{1}{{\sqrt {2t + 1} }}} \right) = \hfill \\ \hfill \\ = \frac{d}{{dt}}{\left( {2t + 1} \right)^{ - 1/2}} = - \frac{1}{2} \cdot {\left( {2t + 1} \right)^{ - \frac{1}{2} - 1}} \cdot 2 = ...? \hfill \\ \end{gathered}$

16. Michele_Laino

correct! Now, you can set t=4, into that last second derivative

17. anonymous

Okay I got -1/27

18. Michele_Laino

that's right!

19. anonymous

Thank you so much for your help!!

20. Michele_Laino

:)

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