Solution: Visibility on Circular Track
1. Relative Motion Analysis
Let the two students be A and B.
- Duration $t = 2 \, \text{min} = 120 \, \text{s}$.
- A completes 3 revolutions. $\omega_A = 3/2 = 1.5 \, \text{rpm}$.
- B completes 4 revolutions. $\omega_B = 4/2 = 2.0 \, \text{rpm}$.
The relative angular velocity is:
$$\omega_{rel} = \omega_B – \omega_A = 2.0 – 1.5 = 0.5 \, \text{rpm}$$In the duration of 2 minutes, the total relative angular displacement is:
$$\Delta \theta_{rel} = \omega_{rel} \times t = 0.5 \times 2 = 1 \, \text{revolution}$$This means that relative to Student A, Student B completes exactly one full circle around the track.
2. Visibility Geometry
The problem states that due to vegetation, “either of the boys can see only one third of the track at a time”.
This implies that from any position on the track, the line of sight allows visibility of an arc corresponding to $1/3$ of the circumference. In terms of angles, the window of visibility is:
$$\theta_{\text{visible}} = \frac{1}{3} \times 360^\circ = 120^\circ$$For the students to see each other, B must be located within this $120^\circ$ angular sector relative to A.
3. Calculation of Time
Since the relative motion is uniform and covers the entire $360^\circ$ of possible relative positions exactly once, the probability (or fraction of time) they are visible is simply the ratio of the visible angle to the total angle.
$$\text{Time Visible} = \text{Total Time} \times \frac{\text{Visible Angle}}{360^\circ}$$ $$\text{Time Visible} = 120 \, \text{s} \times \frac{120^\circ}{360^\circ}$$ $$\text{Time Visible} = 120 \, \text{s} \times \frac{1}{3} = 40 \, \text{s}$$Correct Answer: (b) 40 s