@ganeshie8 @phi
I already found the equation for the velocity and position, if that helps.
can you post them?
Yes. I got \(\Large \sin(\pi t)-\frac{1}{2}\) for the velocity
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
Oh, there's another piece of information. It says x(0) = 0 for the position formula.
we don't know the time, so I would do \[ \int_0^T \sin(\pi t) -\frac{1}{2} \ dt \]
What do you mean?
to get the distance from the origin , integrate velocity from 0 to some time T
your answer for the distance does not look correct, so redo the integral of the velocity
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?
Oh, okay.
Can you check my work to find what I did wrong?
@phi u there?
ok, when integrating sin pi t but -½ dt gives -t/2 term
in other words, you integrate term by term \[ \sin \pi t - \frac{1}{2} \]
So are you saying I'm just missing a -t/2 term? Where did that come from exactly?
your velocity is sin pi t - ½
yes, I know.
So, when I do the u-substitution to find s(t), that part looks okay, right?
yes \[ \int \sin(\pi t) - \frac{1}{2} dt \\ = -\frac{\cos(\pi t)}{\pi} - \frac{t}{2} + c \]
but where does the -t/2 part come from?
if you had just \[ \int \sin(\pi t) \ dt \] you would get \[ -\frac{\cos(\pi t)}{\pi} + c\]
but you have \[ \int \sin(\pi t) - \frac{1}{2} dt \\ \text{ or } \\ \int \sin(\pi t) \ dt- \int \frac{1}{2} dt\]
if you integrate term by term
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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