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anonymous

  • one year ago

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  1. Michele_Laino
    • one year ago
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    is your function like this: \[\Large s\left( t \right) = \sqrt {2t + 1} \]

  2. anonymous
    • one year ago
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    Yes

  3. Michele_Laino
    • one year ago
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    In order to find the acceleration, we have to compute the second derivative of that function s(t)

  4. Michele_Laino
    • one year ago
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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

  5. anonymous
    • one year ago
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    do you guys know what website I can take screen shots with?

  6. anonymous
    • one year ago
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    Michele can you help me with a question???

  7. anonymous
    • one year ago
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    When I derive it I got 1/sqr root(2t+1)

  8. Michele_Laino
    • one year ago
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    correct!

  9. anonymous
    • one year ago
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    Please don't ask questions on my question :)

  10. anonymous
    • one year ago
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    And I would plug in 4 into this?

  11. Michele_Laino
    • one year ago
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    \[\Large \frac{{ds}}{{dt}} = \frac{1}{2} \cdot {\left( {2t + 1} \right)^{\frac{1}{2} - 1}} \cdot 2 = \frac{1}{{\sqrt {2t + 1} }}\]

  12. anonymous
    • one year ago
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    Sorry but Michele can you help me with a question I will fan and medal

  13. Michele_Laino
    • one year ago
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    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
    • one year ago
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    Okay I got -1/(2t+1)^(3/2)

  15. Michele_Laino
    • one year ago
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    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
    • one year ago
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    correct! Now, you can set t=4, into that last second derivative

  17. anonymous
    • one year ago
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    Okay I got -1/27

  18. Michele_Laino
    • one year ago
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    that's right!

  19. anonymous
    • one year ago
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    Thank you so much for your help!!

  20. Michele_Laino
    • one year ago
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    :)

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