Answer
$$
v(t)=s^{\prime}(t)=\sin t-\cos t , \quad s(0)=0
$$
The position of the particle is
$$
s(t)=-\cos t-\sin t+1
$$
Work Step by Step
$$
v(t)=s^{\prime}(t)=\sin t-\cos t , \quad s(0)=0
$$
The general anti-derivative of $
s^{\prime}(t)=\sin t-\cos t
$ is
$$
s(t)=-\cos t-\sin t+C
$$
To determine C we use the fact that $s(0)=0$:
$$
s(0)=-\cos (0)-\sin (0)+C =0
$$
$ \Rightarrow $
$$
-1+C=0 \quad \Rightarrow \quad C=1,
$$
so the position of the particle is
$$
s(t)=-\cos t-\sin t+1.
$$
