Two students simultaneously start from the same place on a circular track and run for 2 min. In this

Q: Why is the visible angle $120^\circ$ and not $60^\circ$?

The problem states they can see one-third of the *entire track*. The track is the circumference of the circle ($360^\circ$). One-third of $360^\circ$ is $120^\circ$. Since the observer is in the middle of this visible arc, it extends $60^\circ$ forward and $60^\circ$ backward.

Q: Why do we use relative angular velocity instead of linear velocity?

Because the track is circular and the visibility condition is based on the angle between the two boys. Angular velocity directly tracks this angle, making the math much simpler than dealing with arc lengths or linear speeds.

Q: What happens when the second boy goes out of sight?

Once the angular separation exceeds $60^\circ$, the line of sight between the two boys intersects the thick vegetation in the center, blocking their view of each other until the faster boy catches up from behind.

Q: Why did we add the two $\pi/3$ intervals at the end?

In one complete relative revolution, the faster boy pulls ahead and is visible for the first $60^\circ$ ($\pi/3$). Then he disappears. He only becomes visible again when he approaches the slower boy from behind, entering the rear $60^\circ$ ($\pi/3$) visible zone.

Q: Would the answer change if the track radius was larger?

No! As long as the vegetation scales proportionally so that exactly one-third of the track is visible, the angles remain identical. The time depends only on the angular speeds, which are determined by the number of revolutions and total time, independent of the radius.

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?

Options:

(A) 20 s

(B) 40 s

(C) 80 s

(D) 160 s

Sigma Percentile · Accepted Answer

### The Geometry of the Hidden Track

Imagine standing on a massive circular track. Right in the center, there is a dense, impenetrable circular patch of vegetation. You look across the field, but your line of sight is blocked. You can only see the parts of the track that your eyes can reach by grazing the edges of this central forest.

The problem gives us a beautiful geometric constraint: at any moment, a runner can only see exactly one-third of the entire track.

Since the total track represents a full $360^\circ$ circle, one-third of this track corresponds to an arc that subtends $120^\circ$ at the center. Because you are standing exactly in the middle of your own field of view, this visible arc extends symmetrically: $60^\circ$ ahead of you, and $60^\circ$ behind you.

Therefore, the fundamental condition for the two boys to see each other is that their angular separation, let's call it $\Delta	heta$, must be less than or equal to $60^\circ$ (or $\frac{\pi}{3}$ radians).

### Analyzing the Runners' Kinematics

Now that we understand the visual boundary, let's look at how fast the boys are moving. Both students run for exactly 2 minutes, which is $120$ seconds.

The first boy completes 3 full revolutions in this time. His angular velocity $\omega_1$ is the total angle covered divided by the time:
$$ \omega_1 = \frac{3 	imes 2\pi}{120} = \frac{\pi}{20} 	ext{ rad/s} $$

The second boy is faster, completing 4 full revolutions. His angular velocity $\omega_2$ is:
$$ \omega_2 = \frac{4 	imes 2\pi}{120} = \frac{\pi}{15} 	ext{ rad/s} $$

### The Elegance of Relative Motion

Tracking two moving objects simultaneously can be mentally taxing. Instead, let's use one of the most powerful tools in physics: **Relative Motion**.

Let's shift our perspective and "sit" on the shoulders of the first boy. In this rotating frame of reference, the first boy is completely stationary. The second boy, however, is moving away from him with a relative angular velocity $\omega_{	ext{rel}}$.

We can calculate this relative speed by simply subtracting their individual speeds:
$$ \omega_{	ext{rel}} = \omega_2 - \omega_1 = \frac{\pi}{15} - \frac{\pi}{20} = \frac{4\pi - 3\pi}{60} = \frac{\pi}{60} 	ext{ rad/s} $$

In this relative frame, the second boy is running laps around the stationary first boy at a speed of $\frac{\pi}{60}$ radians per second.

### The Final Calculation

How long does it take for the second boy to complete exactly one "relative" lap around the first boy? We can find the relative time period $T_{	ext{rel}}$:
$$ T_{	ext{rel}} = \frac{2\pi}{\omega_{	ext{rel}}} = \frac{2\pi}{\pi/60} = 120 	ext{ s} $$

This is a fascinating result! The time it takes to complete one relative revolution is exactly $120$ seconds, which is the total duration of their run. This means the second boy laps the first boy exactly once during the entire event.

During this single relative lap, for how much time is the second boy actually visible?

Remember our geometric constraint: he is visible only when he is within $\frac{\pi}{3}$ radians of the first boy.
This happens in two phases:
1. **The Breakaway:** Right at the start, as the faster boy pulls ahead, he remains visible until he reaches an angle of $\frac{\pi}{3}$ ahead.
2. **The Catch-up:** At the very end of the run, as he completes his lap and approaches the first boy from behind, he enters the visible zone again from $-\frac{\pi}{3}$ (or $\frac{5\pi}{3}$) to $0$.

The total angular distance over which they can see each other is:
$$ 	heta_{	ext{visible}} = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3} 	ext{ rad} $$

Finally, the total time they remain visible is this angular distance divided by their relative speed:
$$ t_{	ext{visible}} = \frac{	heta_{	ext{visible}}}{\omega_{	ext{rel}}} = \frac{2\pi/3}{\pi/60} = 40 	ext{ s} $$

**The boys remain visible to each other for exactly 40 seconds.**

This problem beautifully demonstrates how a complex scenario involving two moving bodies and a geometric constraint can be elegantly unraveled using relative angular kinematics.

Select Answer:

Visualized Solution

\text{Visualizing the Field of View}

\text{Angular Span of Visibility}

\text{Angular Speeds}

\text{Relative Motion Setup}

\text{Relative Time Period}

\text{Calculating Visible Time}

\text{Food for Thought}

The Sigma Insight: Kinematics of Circular Motion

The Geometry of the Hidden Track

Analyzing the Runners' Kinematics

The Elegance of Relative Motion

The Final Calculation

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