1. **State the problem:** We have a particle moving along the x-axis with position function $$x(t) = -\frac{1}{2} \cos t - 3t$$ for $$t \geq 0$$. We need to find the acceleration at time $$t = \frac{\pi}{3}$$. 2. **Recall formulas:** - Velocity $$v(t)$$ is the first derivative of position: $$v(t) = x'(t)$$. - Acceleration $$a(t)$$ is the derivative of velocity or the second derivative of position: $$a(t) = x''(t)$$. 3. **Find velocity:** $$x(t) = -\frac{1}{2} \cos t - 3t$$ Differentiate term-by-term: $$v(t) = x'(t) = -\frac{1}{2} \cdot (-\sin t) - 3 = \frac{1}{2} \sin t - 3$$ 4. **Find acceleration:** Differentiate velocity: $$a(t) = v'(t) = \frac{1}{2} \cos t - 0 = \frac{1}{2} \cos t$$ 5. **Evaluate acceleration at $$t = \frac{\pi}{3}$$:** $$a\left(\frac{\pi}{3}\right) = \frac{1}{2} \cos \frac{\pi}{3} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$$ 6. **Answer:** The acceleration at $$t = \frac{\pi}{3}$$ is $$\frac{1}{4}$$, which corresponds to option (C).