D

Step 1: Start with the given position vector \\( extbf{r}(t) = (4 \cos(2t), 4 \sin(2t), 6t) \\).
Step 2: To find acceleration, first compute the velocity vector by differentiating the position vector: \\n\textbf{v}(t) = \frac{d\textbf{r}(t)}{dt} = \left(\frac{d}{dt}(4 \cos(2t)), \frac{d}{dt}(4 \sin(2t)), \frac{d}{dt}(6t)\right) \
\textbf{v}(t) = (-8 \sin(2t), 8 \cos(2t), 6)
Step 3: Now, compute the acceleration by differentiating the velocity vector: \\n\textbf{a}(t) = \frac{d\textbf{v}(t)}{dt} = \left(\frac{d}{dt}(-8 \sin(2t)), \frac{d}{dt}(8 \cos(2t)), \frac{d}{dt}(6)\right) = (-16 \cos(2t), -16 \sin(2t), 0).
Step 4: Finally, substitute t = \frac{\pi}{2} into the acceleration vector:
At \ t = \frac{\pi}{2}, \ \textbf{a}\left(\frac{\pi}{2}\right) = (-16 \cos(\pi), -16 \sin(\pi), 0) = (16, 0, 0).
Therefore, the acceleration is \textbf{a} = (16, 0, 0) m/s^2.