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Find the position function s(t) from the given velocity or acceleration function and initial value(s). Assume that units are feet and seconds. v(t) = 40 – sin t, s(0) = 2

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@SithsAndGiggles can u help?
\[v(t)=s'(t),\text{ so }\int v(t)~dt+C=s(t)\] So the first thing you do find the indefinite integral of the given \(v(t)\): \[\int(40-\sin t)~dt\]

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ok so s(t) = 40t + cos(t) + c = 40(0) + cos(0) + c = 2
Yeah, that looks right, but when you write it out you have two separate equations: the first being the more general \(s(t)=40t+\cos t+C\) and the second involving the initial values, \(s(0)=40(0)+\cos 0+C=2\). Anyway, solving for C and plugging it into the first equation gives you your answer.
so i should plug in 2 into 40t + cost + c right? and okay
No, you have that \(s(0)=2\), so you would just plug in 0 for t and set the equation equal to 2. You had it right, but you should have written that step in another equation.
ohh ok
so i'm done?

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