1. **State the problem:** We have a glacier whose front edge distance from a car park varies sinusoidally 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 position function:** The glacier's distance $d(t)$ can be modeled as a sine function oscillating between 270 and 280 meters. The amplitude $A$ is half the range: $$A = \frac{280 - 270}{2} = 5$$ The midline (average distance) is: $$D = \frac{280 + 270}{2} = 275$$ The period $T$ is 12 months, so angular frequency $\omega$ is: $$\omega = \frac{2\pi}{T} = \frac{2\pi}{12} = \frac{\pi}{6}$$ Thus, the position function is: $$d(t) = 275 + 5 \sin\left(\frac{\pi}{6} t\right)$$ 3. **Find acceleration function:** Acceleration is the second derivative of position with respect to time $t$: $$v(t) = d'(t) = 5 \cdot \frac{\pi}{6} \cos\left(\frac{\pi}{6} t\right) = \frac{5\pi}{6} \cos\left(\frac{\pi}{6} t\right)$$ $$a(t) = d''(t) = -5 \cdot \left(\frac{\pi}{6}\right)^2 \sin\left(\frac{\pi}{6} t\right) = -5 \cdot \frac{\pi^2}{36} \sin\left(\frac{\pi}{6} t\right) = -\frac{5\pi^2}{36} \sin\left(\frac{\pi}{6} t\right)$$ 4. **Set acceleration threshold:** The geologists camp when $$|a(t)| > \frac{5\pi^2\sqrt{3}}{72}$$ Substitute $a(t)$: $$\left| -\frac{5\pi^2}{36} \sin\left(\frac{\pi}{6} t\right) \right| > \frac{5\pi^2\sqrt{3}}{72}$$ Divide both sides by $\frac{5\pi^2}{36}$ (positive, so inequality direction preserved): $$|\sin\left(\frac{\pi}{6} t\right)| > \frac{5\pi^2\sqrt{3}/72}{5\pi^2/36} = \frac{\sqrt{3}}{2}$$ 5. **Solve inequality for $t$:** We want $$|\sin(x)| > \frac{\sqrt{3}}{2}$$ where $x = \frac{\pi}{6} t$. The sine function exceeds $\frac{\sqrt{3}}{2}$ in intervals around $\frac{\pi}{3}$ and $\frac{2\pi}{3}$ within one period $[0, 2\pi]$: $$\sin(x) > \frac{\sqrt{3}}{2} \text{ for } x \in \left(\frac{\pi}{3}, \frac{2\pi}{3}\right)$$ $$\sin(x) < -\frac{\sqrt{3}}{2} \text{ for } x \in \left(\frac{4\pi}{3}, \frac{5\pi}{3}\right)$$ So $|\sin(x)| > \frac{\sqrt{3}}{2}$ on two intervals each of length $\frac{\pi}{3}$ per period. 6. **Calculate total time per period:** Each interval length is $\frac{\pi}{3}$, so total length per period where $|\sin(x)| > \frac{\sqrt{3}}{2}$ is: $$2 \times \frac{\pi}{3} = \frac{2\pi}{3}$$ Since $x = \frac{\pi}{6} t$, time $t$ corresponding to interval length $\Delta x$ is: $$\Delta t = \frac{6}{\pi} \Delta x$$ Total time per period is: $$\Delta t = \frac{6}{\pi} \times \frac{2\pi}{3} = 4 \text{ months}$$ 7. **Number of periods per year:** The period is 12 months, so one full cycle per year. 8. **Check the claim:** The geologists camp when acceleration exceeds threshold, which is 4 months per year based on above calculation. The claim states they camp between 7 and 8 months per year, but our calculation shows only 4 months. **Conclusion:** The claim is **not reasonable** based on the model and acceleration threshold given. The geologists would camp about 4 months per year, not 7 to 8 months.