1. **State the problem:** We have a glacier's front edge distance from a car park modeled by a sine function oscillating between 270 m and 280 m over a year (12 months). We want to evaluate if the geologists spend between 7 and 8 months at the camp when the absolute acceleration exceeds $\frac{5\pi^2\sqrt{3}}{72}$ m/month$^2$. 2. **Model the distance function:** The glacier distance oscillates between 270 m (minimum) and 280 m (maximum), so the amplitude $A$ is half the range: $$A = \frac{280 - 270}{2} = 5$$ The midline (average distance) is: $$D = \frac{280 + 270}{2} = 275$$ Let $x(t)$ be the distance at time $t$ months. Since the glacier completes one full cycle in 12 months, the angular frequency is: $$\omega = \frac{2\pi}{12} = \frac{\pi}{6}$$ Thus, the position function is: $$x(t) = 275 + 5\sin\left(\frac{\pi}{6}t\right)$$ 3. **Find the acceleration function:** Acceleration is the second derivative of position with respect to time: $$x'(t) = 5 \cdot \frac{\pi}{6} \cos\left(\frac{\pi}{6}t\right) = \frac{5\pi}{6} \cos\left(\frac{\pi}{6}t\right)$$ $$x''(t) = -\frac{5\pi}{6} \cdot \frac{\pi}{6} \sin\left(\frac{\pi}{6}t\right) = -\frac{5\pi^2}{36} \sin\left(\frac{\pi}{6}t\right)$$ 4. **Determine when the absolute acceleration exceeds the threshold:** We want: $$|x''(t)| > \frac{5\pi^2\sqrt{3}}{72}$$ Substitute $x''(t)$: $$\left| -\frac{5\pi^2}{36} \sin\left(\frac{\pi}{6}t\right) \right| > \frac{5\pi^2\sqrt{3}}{72}$$ Simplify by dividing both sides by $\frac{5\pi^2}{36}$ (positive, so inequality direction stays): $$|\sin\left(\frac{\pi}{6}t\right)| > \frac{5\pi^2\sqrt{3}/72}{5\pi^2/36} = \frac{\sqrt{3}}{2}$$ 5. **Solve the inequality for $t$:** We know $|\sin(\theta)| > \frac{\sqrt{3}}{2}$ means $\sin(\theta)$ is in intervals where sine is greater than $\sqrt{3}/2$ or less than $-\sqrt{3}/2$. The critical angles are: $$\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$$ 6. **Find intervals for $\theta = \frac{\pi}{6}t$ where $|\sin(\theta)| > \frac{\sqrt{3}}{2}$:** - For $\sin(\theta) > \frac{\sqrt{3}}{2}$: $$\theta \in \left(\frac{\pi}{3}, \frac{2\pi}{3}\right) + 2k\pi$$ - For $\sin(\theta) < -\frac{\sqrt{3}}{2}$: $$\theta \in \left(\frac{4\pi}{3}, \frac{5\pi}{3}\right) + 2k\pi$$ 7. **Convert $\theta$ back to $t$:** Since $\theta = \frac{\pi}{6}t$, multiply intervals by $\frac{6}{\pi}$: - First interval: $$t \in \left(\frac{\pi}{3} \cdot \frac{6}{\pi}, \frac{2\pi}{3} \cdot \frac{6}{\pi}\right) = (2, 4)$$ - Second interval: $$t \in \left(\frac{4\pi}{3} \cdot \frac{6}{\pi}, \frac{5\pi}{3} \cdot \frac{6}{\pi}\right) = (8, 10)$$ 8. **Length of each interval:** Each interval is $4 - 2 = 2$ months long. 9. **Number of intervals in one year:** Since sine is periodic with period 12, these two intervals occur once per year. 10. **Total time geologists camp:** $$2 + 2 = 4 \text{ months}$$ 11. **Conclusion:** The geologists will camp for 4 months per year, which is less than the claimed 7 to 8 months. **Final answer:** The claim that geologists spend between 7 and 8 months at the camp site is **not reasonable** based on the acceleration threshold and the sine model.