Answer
$s(t)=2\sin t-4\cos t+7$
Work Step by Step
Given : $v(t)=2\cos t+4\sin t$
Recall: The antiderivative of $f(t)=\cos t$ is $F(t)=\sin t$ and the antiderivative of $g(t)=\sin t$ is $G(t)=-\cos t$.
Using this knowledge, the antiderivative of $v(t)$ is $2\sin t+4(-\cos t)+C$.
Since the position formula for a particle is an antiderivative of its velocity, we get $s(t)=2\sin t-4\cos t+C$.
Substituting $s(0)=3$,
$2\sin 0-4\cos 0+C=3$
$2\cdot 0-4\cdot 1+C=3$
$0-4+C=3$
$C=7$
Thus, the position is formulated by $s(t)=2\sin t-4\cos t+7$.
