Kinda hard to tell what [tex]a(t)[/tex] is... In any case, you can find [tex]v(t)[/tex] and [tex]r(t)[/tex] by using the fundamental theorem of calculus:
[tex]v(t)=\displaystyle v(0)+\int_0^ta(u)\,\mathrm du[/tex]
[tex]r(t)=\displaystyle r(0)+\int_0^tv(u)\,\mathrm du[/tex]
If you meant
[tex]a(t)=-\cos t\,\vec\imath-\sin t\,\vec\jmath+3t\,\vec k[/tex]
then we have
[tex]\displaystyle\int_0^t(-\cos u)\,\mathrm du=-\sin t[/tex]
[tex]\displaystyle\int_0^t(-\sin u)\,\mathrm du=\cos t[/tex]
[tex]\displaystyle\int_0^t3u\,\mathrm du=\frac{3t^2}2[/tex]
and so
[tex]v(t)=(\vec\imath+\vec k)+\left(-\sin t\,\vec\imath+\cos t\,\vec\jmath+\dfrac32t^2\,\vec k\right)[/tex]
[tex]v(t)=(1-\sin t)\,\vec\imath+\cos t\,\vec\jmath+\left(1+\dfrac{3t^2}2\right)\,\vec k[/tex]
then
[tex]\displaystyle\int_0^t(1-\sin u)\,\mathrm du=t+\cos t[/tex]
[tex]\displaystyle\int_0^t\cos u\,\mathrm du=\sin t[/tex]
[tex]\displaystyle\int_0^t\frac{3u^2}2\,\mathrm du=\frac{t^3}2[/tex]
so that
[tex]r(t)=(\vec\imath+\vec\jmath+\vec k)+\left((t+\cos t)\,\vec\imath+\sin t\,\vec\jmath+\frac{t^3}2\,\vec k\right)[/tex]
[tex]r(t)=(1+t+\cos t)\,\vec\imath+(1+\sin t)\,\vec\jmath+\dfrac{2+t^3}2\,\vec k[/tex]
