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Mathematics 19 Online
OpenStudy (studygurl14):

@ganeshie8 @phi

OpenStudy (studygurl14):

OpenStudy (studygurl14):

I already found the equation for the velocity and position, if that helps.

OpenStudy (phi):

can you post them?

OpenStudy (studygurl14):

Yes. I got \(\Large \sin(\pi t)-\frac{1}{2}\) for the velocity

OpenStudy (studygurl14):

And I got \(\Large -\frac{1}{2}\cos(\pi t)+\frac{1}{\pi}\) for the position. I'm not sure if this one is correct though

OpenStudy (studygurl14):

Oh, there's another piece of information. It says x(0) = 0 for the position formula.

OpenStudy (phi):

we don't know the time, so I would do \[ \int_0^T \sin(\pi t) -\frac{1}{2} \ dt \]

OpenStudy (studygurl14):

What do you mean?

OpenStudy (phi):

to get the distance from the origin , integrate velocity from 0 to some time T

OpenStudy (phi):

your answer for the distance does not look correct, so redo the integral of the velocity

OpenStudy (studygurl14):

that would be the position formula I got, wouldn't it? So, do I plug in 0 for position and solve for t to find the answer?

OpenStudy (studygurl14):

Oh, okay.

OpenStudy (studygurl14):

Can you check my work to find what I did wrong?

OpenStudy (studygurl14):

@phi u there?

OpenStudy (phi):

ok, when integrating sin pi t but -½ dt gives -t/2 term

OpenStudy (phi):

in other words, you integrate term by term \[ \sin \pi t - \frac{1}{2} \]

OpenStudy (studygurl14):

So are you saying I'm just missing a -t/2 term? Where did that come from exactly?

OpenStudy (phi):

your velocity is sin pi t - ½

OpenStudy (studygurl14):

yes, I know.

OpenStudy (studygurl14):

So, when I do the u-substitution to find s(t), that part looks okay, right?

OpenStudy (phi):

yes \[ \int \sin(\pi t) - \frac{1}{2} dt \\ = -\frac{\cos(\pi t)}{\pi} - \frac{t}{2} + c \]

OpenStudy (studygurl14):

but where does the -t/2 part come from?

OpenStudy (phi):

if you had just \[ \int \sin(\pi t) \ dt \] you would get \[ -\frac{\cos(\pi t)}{\pi} + c\]

OpenStudy (phi):

but you have \[ \int \sin(\pi t) - \frac{1}{2} dt \\ \text{ or } \\ \int \sin(\pi t) \ dt- \int \frac{1}{2} dt\]

OpenStudy (phi):

if you integrate term by term

OpenStudy (phi):

I get \[ d = - \frac{\cos\pi t}{\pi} - \frac{t}{2} + \frac{1}{\pi} \] and we want to find the first t>0 when d =0 I think we have to solve this numerically. Wolfram gets this http://www.wolframalpha.com/input/?i=-1%2Fpi+cos%28pi+*+t%29+-+t%2F2+%2B1%2Fpi%3D0

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