KINEMATICS CYU 17

Step 1: Analyzing Relative Velocities

To simplify the problem, we analyze the motion in the frame of reference of Boy C. Let the direction of motion be the positive x-axis ($\hat{i}$) and the direction perpendicular to the tracks be the y-axis ($\hat{j}$).

Given velocities:

• $\vec{v}_A = 4\hat{i} \text{ m/s}$
• $\vec{v}_B = 2\hat{i} \text{ m/s}$
• $\vec{v}_C = 2\hat{i} \text{ m/s}$

Relative velocities with respect to Boy C:

• $\vec{v}_{A/C} = \vec{v}_A – \vec{v}_C = 4 – 2 = 2\hat{i} \text{ m/s}$
• $\vec{v}_{B/C} = \vec{v}_B – \vec{v}_C = 2 – 2 = 0 \text{ m/s}$

In this frame, Boy C is stationary at the origin $(0,0)$. Boy B is also stationary relative to C. Boy A moves away at $2 \text{ m/s}$. The poles, which are stationary in the ground frame, appear to move backward with velocity $\vec{v}_{P/C} = -2\hat{i} \text{ m/s}$.

Step 2: Geometry of the Field of View

Let the perpendicular distance between the tracks be $d$.

• Position of C: $(0, 0)$
• Position of B (Track 1): $(0, d)$ (Since B is initially collinear with C and has 0 relative velocity)
• Position of A (Track 1): $(2t, d)$ (Since A moves at $2 \text{ m/s}$)

Boy C looks through the space between A and B. We need to find the projection of this gap onto the Power-line track, which is at a distance $2d$ from C (since Track 1 is equidistant from C and the Power-line).

Using similar triangles (intercept theorem):
A line from the origin $(0,0)$ passing through a point $(x, d)$ will intersect the line $y=2d$ at $(2x, 2d)$.

• Left edge of view (Ray passing through B): Passes $(0, d) \rightarrow$ Hits power-line at $x_L = 0$.
• Right edge of view (Ray passing through A): Passes $(2t, d) \rightarrow$ Hits power-line at $x_R = 2(2t) = 4t$.

So, in the frame of C, the visible segment of the power-line at time $t$ is the interval $[0, 4t]$.

Step 3: Pole Visibility Condition

The poles are spaced $D = 18 \text{ m}$ apart. Let the $k$-th pole be initially at position $X_k = 18k$ (where $k = 1, 2, 3 \dots$).
In the frame of C, the poles move to the left at $2 \text{ m/s}$. The position of the $k$-th pole at time $t$ is: $$x’_k(t) = 18k – 2t$$

For the pole to be visible, it must lie within the visible window $[0, 4t]$. $$0 \le x’_k(t) \le 4t$$ $$0 \le 18k – 2t \le 4t$$

This gives us two inequalities:

1. Entrance Condition (Right Inequality):
   $18k – 2t \le 4t \implies 18k \le 6t \implies t \ge 3k$
   The $k$-th pole enters the view at $t = 3k$.
2. Exit Condition (Left Inequality):
   $0 \le 18k – 2t \implies 2t \le 18k \implies t \le 9k$
   The $k$-th pole leaves the view at $t = 9k$.

Thus, the $k$-th pole is visible during the time interval $t \in [3k, 9k]$.

Step 4: Calculating Number of Poles (N) vs Time

We sum the number of poles visible at any given time.

• Pole 1 ($k=1$): Visible for $t \in [3, 9]$.
• Pole 2 ($k=2$): Visible for $t \in [6, 18]$.
• Pole 3 ($k=3$): Visible for $t \in [9, 27]$.
• Pole 4 ($k=4$): Visible for $t \in [12, 36]$.
• Pole 5 ($k=5$): Visible for $t \in [15, 45]$.
• …and so on.

Let’s construct the timeline of events:

Step 5: Graphical Representation

The graph is a step function that increases as new poles enter the field of view and plateaus when the rate of poles entering equals the rate of poles leaving.