1. The problem asks to find the acceleration of the particle at time $t=\pi$ given the position function $x(t) = \sin(2t) - \cos(3t)$ for $t \geq 0$. 2. Acceleration is the second derivative of position with respect to time: $$a(t) = x''(t)$$ 3. First, find the first derivative (velocity): $$x'(t) = \frac{d}{dt}[\sin(2t)] - \frac{d}{dt}[\cos(3t)] = 2\cos(2t) + 3\sin(3t)$$ 4. Next, find the second derivative (acceleration): $$x''(t) = \frac{d}{dt}[2\cos(2t)] + \frac{d}{dt}[3\sin(3t)] = -4\sin(2t) + 9\cos(3t)$$ 5. Evaluate acceleration at $t=\pi$: $$a(\pi) = -4\sin(2\pi) + 9\cos(3\pi)$$ 6. Use known values: $$\sin(2\pi) = 0, \quad \cos(3\pi) = \cos(\pi) = -1$$ 7. Substitute: $$a(\pi) = -4 \times 0 + 9 \times (-1) = -9$$ 8. Therefore, the acceleration at $t=\pi$ is $-9$. The correct answer is E: -9.