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swin2013

A particle moves along a line so that at any time t> (or equal to) 0 its velocity is given by v(t) = t/1+t^2. At time t = 0, the position of the particle is s(0) = 5. Determine the position of the particle at t = 3.

  • one year ago
  • one year ago

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  1. swin2013
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    @zepdrix

    • one year ago
  2. swin2013
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    i know i must take the antiderivative

    • one year ago
  3. zepdrix
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    \[\large s(t)=\int\limits v(t) dt\]Having trouble with the integral? It looks like another nice easy `U substitution`.

    • one year ago
  4. swin2013
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    yea i did u sub for 1 + t^2

    • one year ago
  5. swin2013
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    i meant +

    • one year ago
  6. swin2013
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    i got 1/u du

    • one year ago
  7. zepdrix
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    \[\large \int\limits \frac{1}{u}du\]Yep, sounds good.

    • one year ago
  8. zepdrix
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    Woops, should be a factor of 1/2 in front I think.

    • one year ago
  9. swin2013
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    and i should get u = 1 and u = 10?

    • one year ago
  10. swin2013
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    oh yea 1/2 (1/u) du

    • one year ago
  11. zepdrix
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    No we don't want to add limits to our integral.

    • one year ago
  12. swin2013
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    no?? :O

    • one year ago
  13. zepdrix
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    Evaluate the `Indefinite Integral` to get a function of position, s(t). It will include an unknown constant +C. We can use our `Initial Conditions` to solve for C.

    • one year ago
  14. swin2013
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    oh wait... i do the t(3) = t(0) + the integral

    • one year ago
  15. zepdrix
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    So the integral gave you something like...\[\large s(t)=\frac{1}{2}\ln(1+t^2)+C\]Yes? :O

    • one year ago
  16. swin2013
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    yes!

    • one year ago
  17. zepdrix
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    Then just plug in the `Initial Condition` to solve for C!\[\large s(0)=5 \qquad \qquad \rightarrow \qquad \qquad 5=\frac{1}{2}\ln(1+0^2)+C\]

    • one year ago
  18. swin2013
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    OHHHHH

    • one year ago
  19. swin2013
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    well i did something similar... but i put t(0) -_-

    • one year ago
  20. zepdrix
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    heh :D

    • one year ago
  21. swin2013
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    so i plug 3 for t?

    • one year ago
  22. zepdrix
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    After you've determined the value of C, yes, the final step is asking for s(3). So you would plug in t=3 and solve.

    • one year ago
  23. swin2013
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    i gotsss 3.848

    • one year ago
  24. swin2013
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    idk if that's right.... scary

    • one year ago
  25. zepdrix
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    Hmm I got 6.15

    • one year ago
  26. swin2013
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    :((((((

    • one year ago
  27. zepdrix
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    \[\large C=5\]\[\large s(3)=\frac{1}{2}\ln\left(1+3^2\right)+5\]

    • one year ago
  28. swin2013
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    oh i subt 5 -_-

    • one year ago
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