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A $15 \text{ cm}$ long line of ants starts crawling. A rebel ant at the back of the line steps out and starts marching forward at a higher speed than the line is moving. On reaching the front of the line, it immediately turns around and marches back at the same speed. When it reaches the back of the line it finds that the line of the remaining ants has moved exactly $15 \text{ cm}$. What distance did the rebel ant travel?

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(Large Version)

**Model:**

We model the ants as points on a line. Real ants have length, but that raises some questions, like what to pick as position (head? center?) and how to handle the turn.

Rebel ant marches to top of column with velocity $v=v_r$, while column marches too with $v=v_c$.

The position $s_1$ of the column head ant and of the rebel ant at time $t=t_1$ is:

$$

s_1 = v_c t_1 + L = v_r t_1 \iff \\

(v_r – v_c) t_1 = L \quad (1)

$$

where $L$ is the length of the column.

Rebel marches back at same speed $v=-v_r$, meets column end at same position $s_2$ at $t=t_2$:

$$

s_2 = s_1 – v_r (t_2-t_1) = s_1 + v_c (t_2-t_1) – L \iff \\

(v_r + v_c) (t_2-t_1) = L \quad (2)

$$

The column of ants marched $15$ cm:

$$

15 = v_c t_2 \quad (3)

$$

The column has a length of $15$ cm as well:

$$

L = 15 \quad (4)

$$

**Task:**

Asked for is the distance the rebel ant marched:

$$

d = v_r t_2

$$

**Solution:**

Using equation $(3)$ we get

$$

d = 15 (v_r / v_c)

$$

Solving $(1)$ and $(2)$ for time gives

$$

t_1 = \frac{L}{v_r – v_c} \\

t_2 – t_1 = \frac{L}{v_r + v_c}

$$

and we can use this to express $(3)$ as:

$$

15 =

L v_c \left(\frac{1}{v_r – v_c} + \frac{1}{v_r + v_c} \right)

= \frac{2 L v_r v_c}{v_r^2 – v_c^2} \\

= \frac{2 L (v_r/v_c)}{(v_r/v_c)^2 – 1}

$$

Introducing $x = v_r / v_c$ we get the equation:

$$

15 (x^2 – 1) = 2 L x \iff \\

\left(x – \frac{L}{15}\right)^2 = \frac{L^2+15^2}{15^2} \iff \\

d = 15 x = L \pm \sqrt{L^2 + 15^2} = 15 (1 \pm \sqrt{2})

$$

We further note that $v_r > v_c$ is needed, to have the rebel reach

the top of the column, so we need $x > 1$ and get

$$

d = 15 (1 + \sqrt{2}) = 36.2 \text{ cm}

$$

**Position over Time:**

(Large Version)

The above diagram was calculated for $v_c=5$ cm/s.

Different velocities lead to a different scaling along the $x$-axis.

Shown are the rebel ant $R$ (in red) and the column head $H$ and tail $T$ (in black).

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