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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.

Biology
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i know i must take the antiderivative
\[\large s(t)=\int\limits v(t) dt\]Having trouble with the integral? It looks like another nice easy `U substitution`.

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Other answers:

yea i did u sub for 1 + t^2
i meant +
i got 1/u du
\[\large \int\limits \frac{1}{u}du\]Yep, sounds good.
Woops, should be a factor of 1/2 in front I think.
and i should get u = 1 and u = 10?
oh yea 1/2 (1/u) du
No we don't want to add limits to our integral.
no?? :O
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.
oh wait... i do the t(3) = t(0) + the integral
So the integral gave you something like...\[\large s(t)=\frac{1}{2}\ln(1+t^2)+C\]Yes? :O
yes!
Then just plug in the `Initial Condition` to solve for C!\[\large s(0)=5 \qquad \qquad \rightarrow \qquad \qquad 5=\frac{1}{2}\ln(1+0^2)+C\]
OHHHHH
well i did something similar... but i put t(0) -_-
heh :D
so i plug 3 for t?
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.
i gotsss 3.848
idk if that's right.... scary
Hmm I got 6.15
:((((((
\[\large C=5\]\[\large s(3)=\frac{1}{2}\ln\left(1+3^2\right)+5\]
oh i subt 5 -_-

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