A particle moves along the $x$-axis so that at any time $t\geq 0$, its velocity is given by $v\left(t\right)=\sin\left(2t\right)$. If the position of the particle at time $t = \frac{\pi}{2}$ is $x = 4$, then what is the particle's position at time $t = 0$?
I know that velocity is the derivative of the position function, so I thought I could find the anti-derivative of $v\left(t\right)$ and used the given position to solve for the integration constant and then would have a formula of the position which would allow for me to solve for the $t=0$. However, the answer I obtain is not one of the choices.
$$ x(t) = \int \sin\left(2t\right)\, \mathrm{d}t = -\frac{1}{2}\cos\left(2t\right) + C $$ so $$ 4 = -\frac{1}{2}\cos\left(2\left[\frac{\pi}{2}\right]\right) + C \qquad \Longrightarrow \qquad C = 4 - \frac{1}{2} $$
But $4 - \frac{1}{2}$ is not one of the possible solutions.
Where did I go wrong?
The general procedure, and the evaluation of $C$, are right. We get $s(t)=-\frac{1}{2}\cos 2t+\frac{7}{2}$. Now put $t=0$, and use the fact that $\cos 0=1$. We get $s(0)=3$.