Question

Two students simultaneously start from the same place on a circular track and run for 2 min. In this time, one of them completes three and the other four revolutions. Due to thick vegetation in a circular area as shown in the figure, either of the boys can see only one third of the track at a time. How long during their run do they remain visible to each other?

• A. 20 s
• B. 40 s
• C. 80 s
• D. 160 s

Answer 1

Answer: (B)

40 s

Answer 2

To solve this problem, we need to determine the relative motion of the two students and the conditions under which they are visible to each other.

1. Calculate the angular speeds of the students: Let the total duration of the run be $T = 2 \text{ min} = 120 \text{ s}$.

For student 1: Number of revolutions = 3 Angular displacement = $3 \times 2\pi = 6\pi$ radians Angular speed of student 1, $\omega_1 = \frac{\text{Angular displacement}}{\text{Time}} = \frac{6\pi}{120} = \frac{\pi}{20} \text{ rad/s}$.

For student 2: Number of revolutions = 4 Angular displacement = $4 \times 2\pi = 8\pi$ radians Angular speed of student 2, $\omega_2 = \frac{\text{Angular displacement}}{\text{Time}} = \frac{8\pi}{120} = \frac{\pi}{15} \text{ rad/s}$.

2. Calculate the relative angular speed: Since both students start from the same place and move on a circular track, their relative angular speed is the difference between their individual angular speeds (assuming they move in the same direction, as implied by the problem statement). $\omega_{rel} = \omega_2 - \omega_1 = \frac{\pi}{15} - \frac{\pi}{20}$ To subtract, find a common denominator (60): $\omega_{rel} = \frac{4\pi}{60} - \frac{3\pi}{60} = \frac{\pi}{60} \text{ rad/s}$. This is the rate at which student 2 gains on student 1.

3. Determine the visibility range: Each student can see one-third of the track at a time. A full circular track corresponds to an angle of $2\pi$ radians. So, the angular range of visibility for each student is $\frac{1}{3} \times 2\pi = \frac{2\pi}{3}$ radians. For two students to be visible to each other, their angular separation must be within this range. If one student is at an angle $\theta$, they can see another student if the other student is within the angular interval $[\theta - \frac{\pi}{3}, \theta + \frac{\pi}{3}]$ (considering the visibility range centered around their position). This means the absolute difference in their angular positions, $|\Delta\theta|$, must be less than or equal to $\frac{\pi}{3}$. So, they are visible when $-\frac{\pi}{3} \le \Delta\theta \le \frac{\pi}{3}$ (modulo $2\pi$).

4. Calculate the time for one relative revolution: The time it takes for student 2 to gain one full revolution ($2\pi$ radians) on student 1 is: $T_{rel} = \frac{2\pi}{\omega_{rel}} = \frac{2\pi}{\pi/60} = 2 \times 60 = 120 \text{ s}$.

5. Calculate the total visibility time: The total duration of the run is 120 seconds. This is exactly equal to the time for one relative revolution ($T_{rel}$). During one relative revolution, the relative angular position $\Delta\theta$ changes from $0$ to $2\pi$. They are visible when $\Delta\theta$ is in the range $[-\frac{\pi}{3}, \frac{\pi}{3}]$ (modulo $2\pi$). This corresponds to an angular interval of $\frac{2\pi}{3}$. The fraction of time they are visible during one relative revolution is: Fraction visible = $\frac{\text{Visible angular range}}{\text{Total angular range in one relative revolution}} = \frac{2\pi/3}{2\pi} = \frac{1}{3}$.

Since the total run time is exactly one relative revolution (120 s), the total time they remain visible to each other is: Total visible time = Fraction visible $\times$ Total run time Total visible time = $\frac{1}{3} \times 120 \text{ s} = 40 \text{ s}$.

Alternatively, we can calculate the time intervals: At $t=0$, $\Delta\theta = 0$, they are visible. They remain visible until $\Delta\theta = \frac{\pi}{3}$. Time taken: $t_1 = \frac{\pi/3}{\pi/60} = 20 \text{ s}$. So, they are visible from $t=0$ to $t=20$ s. The relative angle continues to increase. They become visible again when $\Delta\theta$ (modulo $2\pi$) enters the range $[-\frac{\pi}{3}, \frac{\pi}{3}]$ from the other side. This happens when $\Delta\theta$ reaches $2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$ (which is equivalent to $-\frac{\pi}{3}$). Time taken to reach $\frac{5\pi}{3}$: $t_2 = \frac{5\pi/3}{\pi/60} = 100 \text{ s}$. So, from $t=100$ s, they become visible again. They remain visible until $\Delta\theta = 2\pi$ (which is $0$, the starting point), which occurs at $t = 120 \text{ s}$. The duration of this second interval is $120 \text{ s} - 100 \text{ s} = 20 \text{ s}$. Total visible time = $20 \text{ s} + 20 \text{ s} = 40 \text{ s}$.

The final answer is $\boxed{\text{40 s}}$.