$t^*=\dfrac{2a}{3v}$
(ii) Problem Statement
"Three points are located at the vertices of an equilateral triangle whose side equals $a$. They all start moving simultaneously with velocity $v$ constant in modulus, with the first point heading continually for the second, the second for the third, and the third for the first. How soon will the points converge?"
Reference check: the supplied solution says the three points always form an equilateral triangle of shrinking side $l$ and meet at the centre. It finds $-dl/dt=v-v\cos\tfrac{2\pi}{3}=\tfrac{3v}{2}$, so $a=\tfrac32vt$ and $t=\tfrac{2a}{3v}$. This page gets the same result in every frame and method.
(iii) The Big Idea in Plain Words
Three friends stand at the corners of a triangle. Each walks straight at the next friend. All walk at the same speed. The friends never stop turning, so their paths curve in a spiral.
Symbols: $a$ is the starting side. $v$ is each walker's speed. $l$ is the side at time $t$. $t^*$ is the meeting time.
By symmetry the triangle stays equilateral. It only shrinks and turns. So we only track how fast $l$ shrinks. That rate is $\tfrac{3v}{2}$, and it never changes. The spiral is long, but the time is short.
(iv) The Solution Viewed in Each Frame
1. Ground Frame (stationary observer)
Why: look at how the gap between points 1 and 2 changes.
Point 1 moves straight at 2, so it closes the gap at $v$.
Why: point 2 moves toward 3, not away from 1. We need its part along the line.
The angle between its velocity and the line from 1 to 2 is $120^\circ$. It runs away from 1 by $v\cos120^\circ=-\tfrac v2$. So it closes the gap by $\tfrac v2$ more.
Why: add both effects.
$-\dfrac{dl}{dt}=v+\dfrac v2=\dfrac{3v}{2}$. So $l=a-\tfrac32vt$. Set $l=0$: $t^*=\dfrac{2a}{3v}$.
2. Centre-of-Mass Frame (origin at the centre O)
Why: the three velocities have equal size and are $120^\circ$ apart. They add to zero.
$\vec v_1+\vec v_2+\vec v_3=0$. So total momentum is zero for equal masses. The centre does not move. This frame is not accelerating, so there is no pseudo force. Numbers match the ground frame.
Why: each point is at distance $r=l/\sqrt3$ from O. Split its velocity into inward and sideways parts.
Inward part: $v\cos30^\circ=\tfrac{\sqrt3}{2}v$. Sideways part: $\tfrac v2$.
Why: the inward speed is constant, so time is distance over speed.
$t^*=\dfrac{a/\sqrt3}{\sqrt3v/2}=\dfrac{2a}{3v}$.
3. Rotating Frame (turns with the triangle)
Why: in this frame the sideways motion disappears.
Each point only moves straight in toward O at $\tfrac{\sqrt3}{2}v$. The turning rate $\omega=\dfrac{v/2}{r}$ keeps growing as $r$ shrinks. Pseudo forces (centrifugal, Coriolis, Euler) appear, but we do not need them. Same $t^*$.
4. Medium Frame
Does not apply. There is no air or water in this problem.
- Changes: the paths. They are spirals on the ground and straight lines in the rotating frame.
- Same: the time $t^*=\tfrac{2a}{3v}$, the shrink rate $\tfrac{3v}{2}$, and the path length $vt^*=\tfrac{2a}{3}$.
(v) The Solution by Each Method
Method 1: Relative-Motion Method
- Why: in the centre frame every point heads along a line that makes $30^\circ$ with the line to O.
Inward speed is $v\cos30^\circ=\tfrac{\sqrt3}{2}v$. - Why: distance divided by speed gives time.
Start distance: $a/\sqrt3$. So $t^*=\dfrac{a/\sqrt3}{\sqrt3v/2}=\dfrac{2a}{3v}$.
Method 2: Force Method
Does not suit this problem. No forces are given. The speed is fixed by the rule of motion.
Method 3: Energy Method
Does not suit this problem. The speed is constant, so there is no energy change to use.
Method 4: Graph Method
- Why: the side $l$ shrinks at a steady rate, so its graph against $t$ is a straight line.
Start $l=a$. Slope $-\tfrac{3v}{2}$. - Why: the meeting is where the line hits $l=0$.
For $a=6$ m and $v=2$ m/s the slope is $-3$ m/s. Table: $t=0$: 6 m. $t=0.5$: 4.5 m. $t=1$: 3 m. $t=1.5$: 1.5 m. $t=2$ s: 0 m. The bottom graph in the simulation draws this line.
Method 5: Calculus Method
- Why: write the rate of change of the side.
$\dfrac{dl}{dt}=-\left(v-v\cos\dfrac{2\pi}{3}\right)=-\dfrac{3v}{2}$. - Why: integrate from the start to the meeting.
$-\displaystyle\int_a^0dl=\dfrac{3v}{2}\int_0^{t^*}dt\Rightarrow a=\dfrac{3v}{2}t^*$. - Why: for the spiral, the radius is $r=\tfrac{a}{\sqrt3}\left(1-\tfrac{t}{t^*}\right)$ and the turn angle is $\theta=\tfrac{1}{\sqrt3}\ln\tfrac{t^*}{t^*-t}$.
The angle grows without limit as $t\to t^*$. The points circle the centre endlessly but still arrive in finite time. $t^*=\dfrac{2a}{3v}$.
Fastest: Method 1. Most reliable in an exam: Method 5, since it follows the book. All give $t^*=\tfrac{2a}{3v}$.
(vi) Parameter Table
| Symbol | Meaning | Unit | Live value |
|---|
(vii) Common Exam Traps
- Using $v$ as the closing speed of the side. It is $\tfrac{3v}{2}$, because the chased point also moves toward the chaser's line.
- Using $v\cos60^\circ$ with the wrong sign. The angle is $120^\circ$ and the effect adds.
- Using $t=a/v$. That is the time if only one point moved.
- Thinking the path is a straight line. It is a spiral of length $vt^*=\tfrac{2a}{3}$.
- Forgetting that the points meet at the centre, a distance $a/\sqrt3$ from each start.
Verified: the live numbers, three frames and four methods all give $t^*=\tfrac{2a}{3v}$, with $a$ in m, $v$ in m/s and $t$ in s.