Three boys A, B and C decide to walk on straight tracks parallel to a power-line in which poles are

Q: Why is the power line at $y=2d$?

The problem states Track A/B is equidistant from Track C and the power line. If we set Track C at $y=0$ and Track A/B at $y=d$, the power line must be at $y=2d$ to maintain equal spacing.

Q: Why do we use the equations of lines to find the field of view?

Light travels in straight lines. Boy C's line of sight is restricted by the positions of boys A and B. The boundaries of his vision are exactly the straight lines passing through C and B, and C and A.

Q: Why does the graph have 'double-wide' steps?

A new pole enters the view every 3 seconds (when $6t$ crosses a multiple of 18). However, an old pole leaves the view every 9 seconds (when $2t$ crosses a multiple of 18). At $t=9, 18, 27$s, a pole enters and leaves simultaneously, so the total number of visible poles doesn't increase, creating a step that is twice as long!

Q: What happens exactly at $t=3$ seconds?

At exactly $t=3$s, the right boundary of the field of view ($6t$) is exactly at $18$m. The pole is right on the edge of vision. Since the problem asks for poles seen *through* the space, it's an open interval, so the pole is fully visible just after $t=3$.

Q: Can we solve this using relative velocity?

Yes! In the frame of boy C, he is stationary. Boy B is stationary (since $v_B = v_C = 2$ m/s), and boy A moves forward at $2$ m/s. The power line moves backward at $2$ m/s. The geometry remains the same, but the coordinate method is often more robust for finding exact intersection points.

Question

Three boys A, B and C decide to walk on straight tracks parallel to a power-line in which poles are 18 m apart. Boys A and B walk on the same track while C on a different track in the same direction with velocities 4 m/s, 2 m/s and 2 m/s respectively. The track of boys A and B is equidistant from the power line and from the track of the boy C. In the beginning, all the boys and one of the poles are in a line that is perpendicular to the power-line. Draw a graph to show how does number of poles that the boy C can see through the space between boys A and B vary with time.

Sigma Percentile · Accepted Answer

### The Moving Window: A Tale of Relative Motion and Number Theory

Imagine you are Boy C, walking along the bottom track. You look up and see your friends, A and B, walking on the track ahead of you. Because Boy A is walking twice as fast as Boy B, the gap between them is constantly increasing. This gap acts like a moving, expanding window through which you can see the power line in the distance. The question asks us to graph how many poles you can see through this window as time goes on.

This isn't just a kinematics problem; it's a beautiful intersection of geometry and number theory!

### Setting Up the Geometry

Let's anchor this physical situation with a coordinate system. We can place the track of Boy C on the x-axis, so $y = 0$. The problem states that the track of Boys A and B is equidistant from Track C and the power line. If we place Track A/B at $y = d$, the power line must be located at $y = 2d$.

At $t = 0$, everyone starts at the origin $x = 0$. Given their velocities ($v_C = 2$, $v_B = 2$, $v_A = 4$), we can write their positions at any time $t$:
- Boy C: $(2t, 0)$
- Boy B: $(2t, d)$
- Boy A: $(4t, d)$

Notice something interesting? Boys B and C both have an x-coordinate of $2t$. They are always perfectly aligned vertically!

### Defining the Field of View

Boy C's vision is restricted by the positions of B and A. The left boundary of his vision is the straight line passing through C and B. Since they share the same x-coordinate, this is simply the vertical line $x = 2t$. This line intersects the power line ($y = 2d$) exactly at $x = 2t$.

The right boundary of his vision is the line passing through C $(2t, 0)$ and A $(4t, d)$. Let's find its equation. The slope $m$ is:
$$m = \frac{d - 0}{4t - 2t} = \frac{d}{2t}$$

The equation of this line is $y = \frac{d}{2t}(x - 2t)$. To find where this line of sight hits the power line, we substitute $y = 2d$:
$$2d = \frac{d}{2t}(x - 2t)$$
$$4t = x - 2t \implies x = 6t$$

So, at any given time $t$, Boy C can see the segment of the power line strictly between $x = 2t$ and $x = 6t$.

### The Number Theory of Poles

The poles are spaced $18	ext{ m}$ apart, starting from $x = 0$. So, the poles are located at $x = 18n$ for integers $n = 0, 1, 2, \dots$

To find the number of visible poles $N(t)$, we need to count how many multiples of $18$ fall strictly inside the open interval $(2t, 6t)$. Mathematically, we are looking for the number of integers $n$ that satisfy:
$$2t < 18n < 6t$$

Let's analyze when the number of visible poles changes:
1. **A pole enters the view:** This happens when the right boundary expands past a pole. $6t = 18n \implies t = 3n$. So, a new pole enters at $t = 3, 6, 9, 12, 15 \dots$ seconds.
2. **A pole leaves the view:** This happens when the left boundary moves past a pole. $2t = 18n \implies t = 9n$. So, an old pole leaves at $t = 9, 18, 27 \dots$ seconds.

### The Double-Wide Steps

Now, let's construct the graph of $N(t)$:
- For $t \in (0, 3)$, no poles are visible. $N = 0$.
- At $t = 3$, the pole at $18	ext{ m}$ enters. $N = 1$ for $t \in (3, 6)$.
- At $t = 6$, the pole at $36	ext{ m}$ enters. $N = 2$ for $t \in (6, 9)$.

**Here is the catch!** What happens at $t = 9$?
At $t = 9$, the right boundary hits $54	ext{ m}$, so a new pole enters. But at the exact same instant, the left boundary hits $18	ext{ m}$, so the first pole leaves! The net change in visible poles is zero.

Therefore, $N(t)$ remains $2$ for the interval $t \in (9, 12)$.

This phenomenon repeats every $9$ seconds. The graph is a staircase, but the steps at $N = 2, 4, 6 \dots$ are **twice as wide** (lasting $6$ seconds) as the steps at $N = 1, 3, 5 \dots$ (lasting $3$ seconds). This subtle, beautiful detail is what makes this a truly elite kinematics problem!

Visualized Solution

$C oor d ina t e S y s t e m S e t u p$

$P os i t i o n s a t T im e t$

$D e f inin g t h e F i e l d o f V i e w$

$L e f t B o u n d a r y I n t er sec t i o n$

$R i g h t B o u n d a r y I n t er sec t i o n$

$V i s ib l e S e g m e n t o n P o w er L in e$

$C o n d i t i o n f or V i s ib l e P o l es$

$A na l y z in g t h e J u m p s$

$C o n s t r u c t in g t h e G r a p h$

$T h e D o u b l e - W i d e S t e p s$

The Sigma Insight: Relative Velocity

The Moving Window

Setting Up the Geometry

Defining the Field of View

The Number Theory of Poles

The Double-Wide Steps

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Coordinate System Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Positions at Time t

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Defining the Field of View

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Left Boundary Intersection

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Right Boundary Intersection

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visible Segment on Power Line

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Visible Poles

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Jumps

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Constructing the Graph

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Double−Wide Steps