Answer
$s(t)=-\sin t+\cos t+\frac{8}{\pi}t-1$
Work Step by Step
Given: $a(t)=\sin t-\cos t$ with $s(0)=0$ and $s(\pi)=6$
Recall: The velocity of a particle is an antiderivative of its acceleration.
Using this knowledge, the antiderivative of $a(t)$ is $v(t)=-\cos t-\sin t+C$.
Recall: The position of a particle is an antiderivative of its velocity.
Using this knowledge, the antiderivative of $v(t)$ is $s(t)=-\sin t+\cos t+Ct+D$.
Substituting $s(0)=0$, we get
$-\sin 0+\cos 0+C\cdot 0+D=0$
$0+1+0+D=0$
$D=-1$
Now, we have $s(t)=-\sin t+\cos t+Ct-1$.
Substituting $s(\pi)=6$, we get
$-\sin \pi+\cos \pi+C\cdot \pi-1=6$
$-0+(-1)+C\pi -1=6$
$C\pi=8$
$C=\frac{8}{\pi}$
Thus, the position of the particle is $s(t)=-\sin t+\cos t+\frac{8}{\pi}t-1$.
