# 3 Ants going at different speeds, when they will be at the same place Motion Problem

The circumference of a circle has length 90 centimeters, Three points on the circle divide the circle into three equal lengths. Three ants A, B, and C start to crawl clockwise on the circle, with starting from one of the three points. Initially A is ahead of B and B is ahead of C. Ant A crawls 3 centimeters per second, ant V 5 centimeters, and and C 10 centimeters. How long does it take for the three ants to arrive at the same spot for the first time?

I tried making a list and writing down the numbers, but they seem to never be the same. I know the distance formula is d=rt, but I don't know how to use it to solve this problem. Any help? Thanks!

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Hint. Suppose that the ants meet after $t$ seconds, and measure distance around the circle from where C starts. Then A has travelled $3t$ metres, but had a $60$ metre start for a total of $60+3t$ from the initial point. Likewise B will be a distance 30+5t from the initial point. However B may have travelled a number of times around the circle, say $x$ times more than A, and therefore has travelled $90x$ metres further than A. So we have the equation $$60+3t+90x=30+5t\ .$$ See if you can explain by using similar ideas why $$60+3t+90y=10t\ ,$$ where $y$ is the number of times C has "lapped" A.

Now eliminate $t$ from these two equations; then find the smallest possible values of $x$ and $y$, remembering that while $t$ could be any positive number, $x$ and $y$ must be positive integers.

Good luck!

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At the start, C is at location 0, B is at location 30 and A is at location 60. It is easy to see that C will catch up to B in 6 seconds at location 60. And then they will meet again at location 60 every 18 seconds after that. When will A be at location 60? Well, he starts out there and he gets back there every 30 seconds.

So, B and C will be at the right place in 6, 24, 42, 60, 78 etc. seconds while A will be at the right place in 30, 60, 90 etc. seconds. We now see that they meet for the first time after 60 seconds.

To help verify that this is correct you can see that 0 + 10 * 60 = 600; 30 + 5 * 60 = 330; 60 + 3 * 60 = 240. It is easy to see that 600, 330 and 240 all refer to the same spot on the circle. Formally expressed 600 mod 90 = 330 mod 90 = 240 mod 90 = 60.

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First focus on A and B. Since B goes 2 cm/sec faster than A, B first catches A after 15 seconds; and then every 45 seconds thereafter.

Now look at B and C. Since C goes 5 cm/sec faster than B, C first catches B after 6 seconds, and every 18 seconds there after.

So A and B coincide at times $15 + 45n$ seconds for $n = 0, 1, ...$; and B and C coincide at times $6 + 18k$ seconds for $k = 0, 1, 2,...$.

So you need $15 + 45n=6 + 18k$ for nonnegative integer $n,k$.

The smallest solution is $n=1, k=3$. Giving a time of $60$ seconds.

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Suppose, without affecting the result of the problem, that ant $A$ starts at the polar coordinates of angle $0$ on the unit circle, the other ants will be at position $2\frac{\pi}{3}$ and $4\frac{\pi }{3}$. Let's try to find three functions, $f$, $g$, $h$, that takes $t$ (time) as argument, and yields the angle at which ant $A$, $B$, and $C$ will be at the given $t$ moment, respectively. Since the speed is constant, the three functions will be linear.

We know that:

$f(0)=0$

Since the ant $A$ will walk $3$($\frac{90}{30}$) centimeters in $1$ second, it will walk $\frac{1}{30}$ of the circle in $1$ second, so the new angle will decrease by $2\frac{ \pi}{30}$.

$f(1)=-\frac{\pi}{15}$

$f(t)=at+b$

$f(0)=b=0$

$f(1)=a=-\frac{\pi}{15}=>a=-\frac{\pi}{15}$

Similarly, we can compute for $g$ and $h$.

$f(t)=-\frac{\pi}{15}t$

$g(t)=-\frac{\pi}{9}t+2\frac{\pi}{3}$

$h(t)=-2\frac{\pi}{9}t+4\frac{\pi}{3}$

Now, it only remains to solve the equation(for any $n$, $m$).

$f(t)=g(t)+2n\pi =h(t)+2m\pi$

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Expressing the position by the angle doesn't make anything simpler and fills your calculation with fractions of $\pi$, making it more likely that you'll make mistakes. –  David Richerby Aug 9 '14 at 10:55

Another way of simplifying the problem:

A is at 4 o'clock, B is at 12 o'clock, and C is at 8 o'clock.

Let's subtract $5$ cm/s from each ant's speed. So now C is going clockwise at only $5$ cm/s, B is stationary, and A is going counter-clockwise at $2$ cm/s. And they're all still $30$ cm apart. Subtracting the same amount from each ant's speed doesn't change the solution to the problem: in physics terms, it's equivalent to working in ant B's frame of reference, instead of the circle's.

C will reach B after $\frac{30}{5}$ sec, and thereafter every $\frac{90}{5}$ sec, or after $6, 24, 42, 60, 78$, ... seconds

A will reach B after $\frac{30}{2}$ sec, and thereafter every $\frac{90}{2}$ sec, or after $15, 60, 105$, ... seconds

The first double hit is after 60 seconds...

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I suggest you use the angle as the ant's position indicator. Note that $$\theta = \frac{s}{r} = \frac{s}{{\frac{{90}}{{2\pi }}}} = \frac{{2\pi s}}{{90}}$$Now:$$\begin{array}{l}{\theta _{\rm{A}}}(t) = {\omega _{\rm{A}}}t + {\theta _{{{\rm{A}}_0}}}\\{\theta _{\rm{B}}}(t) = {\omega _{\rm{B}}}t + {\theta _{{{\rm{B}}_0}}}\\{\theta _{\rm{C}}}(t) = {\omega _{\rm{C}}}t + {\theta _{{{\rm{C}}_0}}}\end{array}$$and$$\begin{array}{l}{\theta _{{{\rm{A}}_0}}} = - \frac{{4\pi }}{3}\\{\theta _{{{\rm{B}}_0}}} = - \frac{{2\pi }}{3}\\{\theta _{{{\rm{C}}_0}}} = 0\end{array}$$where$$\begin{array}{l}{\omega _{\rm{A}}} = \frac{{6\pi }}{{90}}\\{\omega _{\rm{B}}} = \frac{{10\pi }}{{90}}\\{\omega _{\rm{C}}} = \frac{{20\pi }}{{90}}\end{array}$$Your question could be re-written as find a $t$ such that$$\begin{array}{l}{\theta _{\rm{C}}}(t) - {\theta _{\rm{B}}}(t)\mathop \equiv \limits^{2\pi } 0\\{\theta _{\rm{C}}}(t) - {\theta _{\rm{A}}}(t)\mathop \equiv \limits^{2\pi } 0\end{array}$$Replacing the data and simplifying yields$$\begin{array}{l}\frac{{\pi t + 6\pi }}{9}\mathop \equiv \limits^{2\pi } 0\\\frac{{7\pi t + 60\pi}}{{45}}\mathop \equiv \limits^{2\pi } 0\end{array}$$ which is equivalent to $$\begin{array}{l}t + 6\mathop \equiv \limits^{18} 0\\7t + 60\mathop \equiv \limits^{90} 0\end{array}$$Combining the second equation with the first, yields $$\begin{array}{l}t\mathop \equiv \limits^{18} 12\\t\mathop \equiv \limits^{30} 0\end{array}$$The first solution to these is $t=30$.

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That makes it seems harder though...what's the the greek alphavet for? I still gave an upvote for answering though. –  Anonymous Aug 9 '14 at 4:04
Expressing the position by the angle doesn't make anything simpler and fills your calculation with fractions of $\pi$, making it more likely that you'll make mistakes. –  David Richerby Aug 9 '14 at 10:55
@Anonymous Using $\theta$ for an angle and $\omega$ for an angular velocity is as standard as using $t$ for time or $r$ for radius. –  David Richerby Aug 9 '14 at 12:56