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swin2013

  • 3 years ago

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.

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  1. swin2013
    • 3 years ago
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    @zepdrix

  2. swin2013
    • 3 years ago
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    i know i must take the antiderivative

  3. zepdrix
    • 3 years ago
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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`.

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

  5. swin2013
    • 3 years ago
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    i meant +

  6. swin2013
    • 3 years ago
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    i got 1/u du

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

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

  9. swin2013
    • 3 years ago
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    and i should get u = 1 and u = 10?

  10. swin2013
    • 3 years ago
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    oh yea 1/2 (1/u) du

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

  12. swin2013
    • 3 years ago
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    no?? :O

  13. zepdrix
    • 3 years ago
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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.

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

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

  16. swin2013
    • 3 years ago
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    yes!

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

  18. swin2013
    • 3 years ago
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    OHHHHH

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

  20. zepdrix
    • 3 years ago
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    heh :D

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

  22. zepdrix
    • 3 years ago
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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.

  23. swin2013
    • 3 years ago
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    i gotsss 3.848

  24. swin2013
    • 3 years ago
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    idk if that's right.... scary

  25. zepdrix
    • 3 years ago
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    Hmm I got 6.15

  26. swin2013
    • 3 years ago
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    :((((((

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

  28. swin2013
    • 3 years ago
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    oh i subt 5 -_-

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