## Calculus 8th Edition

$$a(t)=v^{\prime}(t)= 3 \cos t-2 \sin t, \quad s(0)=0 , \quad v(0)=4$$ The position of the particle is $$s(t)=-3 \cos t+ 2 \sin t +2t +6$$
$$a(t)=v^{\prime}(t)= 3 \cos t-2 \sin t, \quad s(0)=0 , \quad v(0)=4$$ The general anti-derivative of $v^{\prime}(t)= 3 \cos t-2 \sin t$ is $$v(t)= 3 \sin t+ 2 \cos t +C$$ To determine C we use the fact that $v(0)=-2$: $$v(0)= 3 \sin (0)+ 2 \cos (0) +C =4$$ $\Rightarrow$ $$2+C=4 \quad \Rightarrow \quad C=2,$$ so $$v(t)= 3 \sin t+ 2 \cos t +2$$ Since $s^{\prime }(t)=v(t)$, we antidifferentiate again and obtain $$s(t)=-3 \cos t+ 2 \sin t +2t +D$$ To determine D we use the fact that $s(0)=0$: $$s(0)=-3 \cos (0)+ 2 \sin (0) +2(0) +D =3$$ $\Rightarrow$ $$-3+D=3 \quad \Rightarrow \quad D=6,$$ so the position of the particle is $$s(t)=-3 \cos t+ 2 \sin t +2t +6$$