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 one year 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.
 one year 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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swin2013
 one year ago
Best ResponseYou've already chosen the best response.0i know i must take the antiderivative

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0\[\large s(t)=\int\limits v(t) dt\]Having trouble with the integral? It looks like another nice easy `U substitution`.

swin2013
 one year ago
Best ResponseYou've already chosen the best response.0yea i did u sub for 1 + t^2

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0\[\large \int\limits \frac{1}{u}du\]Yep, sounds good.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Woops, should be a factor of 1/2 in front I think.

swin2013
 one year ago
Best ResponseYou've already chosen the best response.0and i should get u = 1 and u = 10?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0No we don't want to add limits to our integral.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Evaluate 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.

swin2013
 one year ago
Best ResponseYou've already chosen the best response.0oh wait... i do the t(3) = t(0) + the integral

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0So the integral gave you something like...\[\large s(t)=\frac{1}{2}\ln(1+t^2)+C\]Yes? :O

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Then just plug in the `Initial Condition` to solve for C!\[\large s(0)=5 \qquad \qquad \rightarrow \qquad \qquad 5=\frac{1}{2}\ln(1+0^2)+C\]

swin2013
 one year ago
Best ResponseYou've already chosen the best response.0well i did something similar... but i put t(0) _

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0After 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.

swin2013
 one year ago
Best ResponseYou've already chosen the best response.0idk if that's right.... scary

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0\[\large C=5\]\[\large s(3)=\frac{1}{2}\ln\left(1+3^2\right)+5\]
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