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0mega

  • 3 years ago

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

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  1. zzr0ck3r
    • 3 years ago
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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

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

  3. zzr0ck3r
    • 3 years ago
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    yerp

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

  5. zzr0ck3r
    • 3 years ago
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    yep

  6. zzr0ck3r
    • 3 years ago
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    note you need to solve for c first

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

  8. 3psilon
    • 3 years ago
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    anti derivative

  9. zzr0ck3r
    • 3 years ago
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    the integral of Bt

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

  11. 0mega
    • 3 years ago
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    ohhhhhh. okay okay. i get it now @zzr0ck3r

  12. zzr0ck3r
    • 3 years ago
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    great... note: Newton was a p i m p

  13. 3psilon
    • 3 years ago
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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

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