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anonymous

  • 4 years ago

What is the acceleration at time t=3 of a particle whose position at time t is given by s(t)=3sqrt(x+1)^(1/3) Note: The square root is the third root and it is being multiplied by another 3 that's the 3 in front.. so, it's 3 * third root(x+1)

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  1. anonymous
    • 4 years ago
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    instead of x, it is t?

  2. anonymous
    • 4 years ago
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    \[s(3)=3(\sqrt[3]{x+1})^{1/3}=3\sqrt[9]{4}\]

  3. phi
    • 4 years ago
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    \[s(t)= 3(t+1)^{\frac{1}{3}}\]

  4. anonymous
    • 4 years ago
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    cube root of a cube root? or what?

  5. anonymous
    • 4 years ago
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    it looks like rickjbr but without the 1/3... sorry

  6. anonymous
    • 4 years ago
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    \[3\sqrt[3]{4}\]

  7. anonymous
    • 4 years ago
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    3+1=4 lol

  8. phi
    • 4 years ago
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    replace the cube root with the equivalent to the 1/3 power. take the 2nd derivative of s(t) to find the acceleration

  9. anonymous
    • 4 years ago
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    oh that is the position...i'm retarded

  10. anonymous
    • 4 years ago
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    is the answer -0.333?

  11. phi
    • 4 years ago
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    you mean (t+1) not 4. plug in t=3 after the 2nd derivative

  12. anonymous
    • 4 years ago
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    whoops, i'm not thinking clearly, phi is right.

  13. phi
    • 4 years ago
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    you are in a rush to get the answer. Must be close to quitting time.

  14. anonymous
    • 4 years ago
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    Yup

  15. anonymous
    • 4 years ago
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    First derivative is (t+1)^(-2/3)

  16. anonymous
    • 4 years ago
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    2nd would be \[\frac{2(t+1)^{-5/3}}{3}\]

  17. phi
    • 4 years ago
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    with a minus sign

  18. anonymous
    • 4 years ago
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    with a minus sign lol, make it negative

  19. anonymous
    • 4 years ago
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    \[\frac{-2}{3(t+1)^{5/3}}\]

  20. anonymous
    • 4 years ago
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    Now you can plug in 3

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