Find the position vector for a particle with acceleration, initial velocity, and initial position given below.
\[
\begin{array}{l}
\vec{a}(t)=\langle 3 t, 4 \sin (t), \cos (5 t)\rangle \\
\vec{v}(0)=\langle 3,3,-4\rangle \\
\vec{r}(0)=\langle 0,-2,3\rangle \\
\vec{r}(t)=
\end{array}
\]
I

Step 1 ：Given the acceleration vector \(\vec{a}(t)=\langle 3 t, 4 \sin (t), \cos (5 t)\rangle\), the initial velocity vector \(\vec{v}(0)=\langle 3,3,-4\rangle\), and the initial position vector \(\vec{r}(0)=\langle 0,-2,3\rangle\)

Step 2 ：The position vector of a particle can be found by integrating the acceleration vector twice. The first integration will give us the velocity vector and the second integration will give us the position vector. We also need to use the initial velocity and position vectors as the constants of integration.

Step 3 ：Integrating the acceleration vector \(\vec{a}(t)=\langle 3 t, 4 \sin (t), \cos (5 t)\rangle\) gives us the velocity vector \(\vec{v}(t)=\langle \frac{3}{2}t^2 + 3, -4\cos(t) + 3, \frac{1}{5}\sin(5t) - 4\rangle\)

Step 4 ：Integrating the velocity vector \(\vec{v}(t)=\langle \frac{3}{2}t^2 + 3, -4\cos(t) + 3, \frac{1}{5}\sin(5t) - 4\rangle\) gives us the position vector \(\vec{r}(t)=\langle \frac{1}{2}t^3 + 3t, 3t - 4\sin(t) - 2, -4t - \frac{1}{25}\cos(5t) + 3 \rangle\)

Step 5 ：\(\boxed{\vec{r}(t)=\langle \frac{1}{2}t^3 + 3t, 3t - 4\sin(t) - 2, -4t - \frac{1}{25}\cos(5t) + 3 \rangle}\) is the final position vector for the particle.