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 $t=0$.

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