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anonymous

  • one year ago

Find the instantaneous velocity and speed at time 5 secs when v(t) = t2 – 9t + 18 and s(0) = 1

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  1. anonymous
    • one year ago
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    i figured i would just ave to find the acceleration (which is just the derivative of v(t)) and then plug five into that but i did that and got it wrong

  2. Astrophysics
    • one year ago
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    Why would you find acceleration when it's asking for velocity, just plug in 5 for your velocity.

  3. anonymous
    • one year ago
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    its asking for the instantaneous velocity doesn't that mean the tangent?

  4. Astrophysics
    • one year ago
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    Are you sure it's not suppose to be s(t)

  5. anonymous
    • one year ago
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    no i am given v(t)

  6. anonymous
    • one year ago
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    but i am also given s(0) so i can just find s(t)

  7. Astrophysics
    • one year ago
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    You're given the instantaneous velocity

  8. Astrophysics
    • one year ago
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    Just find v(5)

  9. anonymous
    • one year ago
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    okay so then v(5) = -2

  10. Astrophysics
    • one year ago
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    Yes

  11. anonymous
    • one year ago
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    okay and for the speed i just find s(t) and plug five into that right?

  12. Astrophysics
    • one year ago
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    s(t) is the position

  13. Astrophysics
    • one year ago
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    speed is just scalar

  14. anonymous
    • one year ago
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    okay so then speet(t)=2 ?

  15. Astrophysics
    • one year ago
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    Yes, speed is just the magnitude or the velocity

  16. anonymous
    • one year ago
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    okay lovely ! thank you very much!

  17. Astrophysics
    • one year ago
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    Yw :)

  18. Astrophysics
    • one year ago
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    We could find the position to s(t), by integrating, \[v(t) = t^2-9t+18 \implies \frac{ ds }{ dt } = t^2-9t+18\]\[\int\limits ds = \int\limits (t^2-9t+18) dt = s(t) = \frac{ t^3 }{ 3 }-\frac{ 9t^2 }{ 2 }+18t+C \]

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