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?
Asked
Active
Viewed 1,212 times
-4
-
2Welcome to Stackexchange. You'll find that simple "Here's the statement of my exercise, solve it for me" posts will be poorly received. What is better is for you to add context: What you understand about the problem, what you've tried so far, etc. Something to both show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – David Jan 06 '16 at 04:51
-
2Assume (it doesn't matter) that the column has speed $1$. Let $r$ be the speed of the rebel ant. Then the time taken by the ant is $\frac{15}{r-1}+\frac{15}{r+1}$. This is also $\frac{15}{1}$. – André Nicolas Jan 06 '16 at 06:13
-
@user302943 Please check and comment my answer. – mvw Jan 06 '16 at 11:52
-
@superbjhon I added more fancy graphics and cleaned up the timing. – mvw Jan 10 '16 at 02:08
-
@AndréNicolas Is this method of letting speed=$1$ exactly correct?Can we let things as $1$ in this way?I have always known we let things as $x$ or $y$... – Soham Mar 10 '16 at 16:33
-
@tatan: Yes, it is a little trick that is correct. Whatever the column's speed is, we can choose the unit of time so that this speed is $1$ cm per time unit. The idea is analogous to the trick one often uses in analytic geometry, when we say, for example, that without loss of generality we may assume that the circle has radius $1$. – André Nicolas Mar 10 '16 at 16:42
-
@AndréNicolas You are great...Thanks!! – Soham Mar 10 '16 at 16:44
-
Of course we can let the speed of the ant be $v_a$, and the column speed $v_c$, and set up equations. But the trick substantially simplifies the calculation. – André Nicolas Mar 10 '16 at 16:46
1 Answers
2
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:
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).
mvw
- 34,562