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snaef999

  • 3 years ago

A particle starts at the origin and initial verlocity is i - j + 3k. Its acceleration is a(t) = 6t i + 12t^2 j + 6t k. Find its position function. I need help confirming my answer. Is the correct way to do this to integrate a(t) to get velocity and then integrate velocity to get position?

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  1. eseidl
    • 3 years ago
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    yeah. You are also given it's initial velocity

  2. snaef999
    • 3 years ago
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    When finding velocity I need to use the initial velocity as the C right?

  3. eseidl
    • 3 years ago
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    Yes, the constant of integration after the first integration is your initial velocity

  4. snaef999
    • 3 years ago
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    My final answer is (t^3 + t)i + (t^4 -t)j + (2/3 t^3 + 3t)k + c (c = 0 because we started at the origin)

  5. eseidl
    • 3 years ago
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    hang on...i'll check and see what I get

  6. eseidl
    • 3 years ago
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    I get s(t)=(t^3+t)i+(t^4-t)j+(t^3-3t)k

  7. snaef999
    • 3 years ago
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    Ah. I see what I did wrong. Thank you for your help! I looked at the wrong numbers for k when I was dong position integral. It works for me now.

  8. eseidl
    • 3 years ago
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    cool, no prob

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