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0mega

a(t) = A + Bt x(0) =xi v(0) = vi derive the position as a function of time equation

  • one year ago
  • one year ago

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  1. zzr0ck3r
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    take the first integral v(t) = At+(1/2)Bt^2 + C then plug in v(0) to solve for C

    • one year ago
  2. 3psilon
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    Then take the next integral and plug X(0)

    • one year ago
  3. zzr0ck3r
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    yerp

    • one year ago
  4. 0mega
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    i take the integral of the first integral? and then plug in x(0)?

    • one year ago
  5. zzr0ck3r
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    yep

    • one year ago
  6. zzr0ck3r
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    note you need to solve for c first

    • one year ago
  7. 0mega
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    how did you get (1/2)bt^2 ?

    • one year ago
  8. 3psilon
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    anti derivative

    • one year ago
  9. zzr0ck3r
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    the integral of Bt

    • one year ago
  10. zzr0ck3r
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    A*0+(1/2)B*0^2 + C = vo c = v0

    • one year ago
  11. 0mega
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    ohhhhhh. okay okay. i get it now @zzr0ck3r

    • one year ago
  12. zzr0ck3r
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    great... note: Newton was a p i m p

    • one year ago
  13. 3psilon
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    \[V(0) = V_{i} \] according to the boundary condition So \[V(0) =\frac{ B0^{2} }{ 2 } + a(0) + C\] \[ V(0) = C\] \[ V(0) = vi\] So \[ Vi = C\] Plug \[ V(i)\] into the orginal \[V(t)\] function as C

    • one year ago
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