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anonymous
 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.
anonymous
 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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i know i must take the antiderivative

zepdrix
 3 years 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`.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yea i did u sub for 1 + t^2

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and i should get u = 1 and u = 10?

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

zepdrix
 3 years 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.

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well i did something similar... but i put t(0) _

zepdrix
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0idk if that's right.... scary

zepdrix
 3 years 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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