A particle moves along the \(x\)-axis so that at any time \(t \gt 0\), its velocity is given by \(v(t)=\sin(2t)\). If the position of the particle at time \(t=\frac{\pi}{2}\) is \(x=4\), what is the particle's position at time \(t=0\) ?

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Integrating the velocity equation with respect to time gives us a position equation:

$$ \int v(t)   dt = \int \sin(2t)   dt $$ $$ s(t) = -\frac{1}{2}\cos(2t)+C$$

It is given that the position of the particle at time \(t=\frac{\pi}{2}\) is \(x=4\).

$$ s\left(\frac{\pi}{2}\right)=4 $$ $$ 4 = -\frac{1}{2}\cos(2\left(\frac{\pi}{2}\right))+C $$ $$ 4 = -\frac{1}{2}\cos(\pi)+C $$ $$ 4 = \frac{1}{2}+C$$ $$ C = \frac{7}{2} $$

Our position equation is therefore:

$$ s(t)=-\frac{1}{2}\cos(2\left(\frac{\pi}{2}\right))+\frac{7}{2} $$

Finding the position at \(t=0\).

$$ s(0)=-\frac{1}{2}\cos(2(0))+\frac{7}{2} $$ $$ s(0)=-\frac{1}{2}+\frac{7}{2} $$ $$ s(0)= \boxed{3} $$