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Denebel

  • 2 years ago

an automobile accelerates from rest at 1+ 3* sqrt (t) mph/sec for 9 seconds. What is the velocity after 9 seconds?

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  1. 14yamaka
    • 2 years ago
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    1+3sqrt(9) 1+3(3) 1+9 10mph

  2. Xishem
    • 2 years ago
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    ^ Wouldn't this find the acceleration at time t, not the velocity?

  3. Xishem
    • 2 years ago
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    *time 9s

  4. 14yamaka
    • 2 years ago
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    Good point Xishem... I wondered about that too, but I can't tell from the question X(

  5. Xishem
    • 2 years ago
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    Would integrating the function from 0 to 9 give the correct answer?

  6. mattt9
    • 2 years ago
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    yes

  7. Denebel
    • 2 years ago
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    Hm, when the problem states it like that... does it mean that the velocity is the function 1 + 3*(t) or the acceleration?

  8. mattt9
    • 2 years ago
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    the function is of acceleration

  9. mattt9
    • 2 years ago
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    so its integral will be velocity

  10. 14yamaka
    • 2 years ago
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    It means acceleration....yeah, Xishem, you're right, this will require integration

  11. Denebel
    • 2 years ago
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    Yeah, the book says mph/sec.. how should it be stated if it was giving the velocity?

  12. mattt9
    • 2 years ago
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    (t+6t^(3/2)) from 0-->9

  13. mattt9
    • 2 years ago
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    i would change either the hours to seconds or vice versa

  14. mattt9
    • 2 years ago
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    so your answer can be in the form mps or mph

  15. mattt9
    • 2 years ago
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    this can be done by dimensional analysis

  16. dumbcow
    • 2 years ago
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    i get v(t) = t + 2t*sqrt(t) v(9) = 9 +2*9*sqrt(9) = 63 mph

  17. Denebel
    • 2 years ago
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    integral of sqrt (t) = [2t^(3/2)]/3 Multiply that by the 3 constant.. = 2t^(3/2) am I doing something wrong? I can't get sqrt (t)

  18. mattt9
    • 2 years ago
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    it's the same thing... 2t*t^(1/2)=2t^(3/2)

  19. Denebel
    • 2 years ago
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    Oh I see. I would have never though of to do that... makes calculations so much easier, haha.

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