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A woman is walking home with distance $d$ and speed $v$.

The dog is happy and runs at speed $\frac{3v }{ 2}$ always between woman and House back and forth.

(i) At what distances $(d_n)_{n\geq 1}$ do women and dogs meet?

(ii) Determine the total path length of the dog with the help of $(d_n)_{n\geq 1}$.

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EDIT:

Let $x$ be the distance between the woman and the house and let $s$ be the distance between the dog and the house.

At the beginning we have $x_0=d$ and $s_0=0$.

It holds that $x=d-vt$ and $s=\frac 32v t$.

The woman and the dog meet at $x=s=d_1$ for $t=t_1$.

From $x=s$ we get $d-vt=\frac{3}{2}vt \Rightarrow \frac{5}{2}vt=d \Rightarrow t=\frac{2d}{5v}$, so $t_1=\frac{2d}{5v}$.

So we have that $d_1=x(t_1)=d-v\cdot \frac{2d}{5v}=d- \frac{2d}{5}=\frac{3d}{5}$.

After the meeting of the woman and the dog, the dog goes back to the house and then again to the woman.

At time $t_1$ the dog is at $s_1=\frac{3d}{5}$.

The dog goes in direction of the house, so $s=\frac{3d}{5}-\frac{3v}{2}t$.

When $s=0$ the dog arrived at the house, so $t=\frac{2d}{5v}$.

Then the dog goes back in direction of the woman, so $s= \frac{3}{2}vt$ with $t>\frac{2d}{5v}$.

At time $t_1$ the woman is also at $x_1=\frac{3d}{5}$.

The woman continues in diection to the house, so $x=\frac{3d}{5}-vt$.

For the second meeting, it holds $x=s=d_2$.

We get \begin{equation*}\frac{3d}{5}-vt=\frac{3}{2}vt \Rightarrow \frac{3d}{5}=\frac{5}{2}vt \Rightarrow t=\frac{6d}{25v}\end{equation*} So they meet at \begin{equation*}d_2=x(t_2)=\frac{3d}{5}-v\cdot \frac{6d}{25v}=\frac{9d}{25}\end{equation*}.

So the $n$-th meeting is at $d_n =\left (\frac{3}{5}\right )^nd$.

Is that correct so far?

For the second question do we take the sum of all the terms $d_n$ ? Or should we take twice each term $d_n$ since the dog goes back and forth?

Mary Star
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    Do the first few steps (e.g., $n=1,2,3$) by hand and look for a pattern. – Semiclassical Jan 19 '21 at 20:27
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    This is reminiscent of the (apocryphal?) story of von Neumann, the fly, and the colliding trains. See https://math.stackexchange.com/questions/312840/infinite-series-anecdote-about-john-von-neumann – Barry Cipra Jan 19 '21 at 20:32
  • Could you please have a look at the edit part of my question above? @Semiclassical – Mary Star Jan 20 '21 at 11:10
  • Could you please have a look at the edit part of my question above? @BarryCipra – Mary Star Jan 20 '21 at 11:10
  • If the woman and the dog start off at distance $d_0$ from the house and meet again when the woman is at distance $d_1$ from the house, the dog will have walked a distance $d_0+d_1$. If they meet again at distance $d_2$, the dog will have walked another distance $d_1+d_2$. So the total distance the dog walks is $$(d_0+d_1)+(d_1+d_2)+\cdots=d_0+2d_1+2d_2+2d_3+\cdots$$ They key thing to notice is that the dog walks the initial distance $d_0$ only once, but all other distances twice. – Barry Cipra Jan 20 '21 at 14:45
  • What I have done for the first question is all correct? @BarryCipra – Mary Star Jan 20 '21 at 17:33

2 Answers2

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The distance of the woman from home is given by $d-vt$, and that of the dog by $d-\frac32vt$, until it reaches it, at time $t_1=\frac23\frac dv$. It comes back to its mistress with equation

$$\frac32v(t-t_1)=\frac32vt-d.$$

The moment they meet is obtained by solving

$$\frac32vt_2-d=d-vt_2$$ or $$t_2=\frac25\frac dv,$$ at a distance $\frac35d$ of home.

Now this repeats proportionally, substituting $\frac35d$ for $d$, and we have a geometric progression of common ratio $\frac35$.

The distances of the meetings are given by the partial sums of the sequence

$$\frac25d,\frac25\frac35d,\frac25\frac9{25}d,\cdots\frac25\frac{3^n}{5^n}d,\cdots$$

which indeed converges to $d$.

The dog travels $\frac32$ times more than the woman, hence $\frac32d$.

  • So my approach is correct just the formulation could be improved, right? – Mary Star Jan 20 '21 at 17:32
  • @MaryStar: correct. –  Jan 20 '21 at 17:36
  • @YvesDaoust, something here seems off to me. Suppose the woman and dog start five miles from home. You seem to be saying that the dog first returns when the woman is three miles from home. But that means the dog has run eight miles (five miles home and three miles back) while the woman has walked only two, which means the dog is going four times faster than the woman, not a mere fifty percent faster. It seems to me that we should get $d_1={1\over5}d$, not ${3\over5}d$. Am I missing something? – Barry Cipra Jan 20 '21 at 18:18
  • When the dog comes back to the woman why is the equation equal to $\frac32v(t-t_1)=\frac32vt-d$ ? I got stuck right now. – Mary Star Jan 20 '21 at 18:18
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If the woman and dog start at distance $d$ from the house and re-meet when the woman is at distance $\delta$ from the house, the woman will have walked a distance $d-\delta$ and the dog will have run a distance $d+\delta$. Now since the dog runs fifty percent faster than the woman walks, it goes fifty percent further in any given amount of time, meaning

$$d+\delta={3\over2}(d-\delta)$$

and thus $\delta={1\over5}d$. So the sequence of distances (from the house) that the woman and the dog meet up is $d_n=\left(1\over5\right)^nd$. The total distance the dog runs is

$$\begin{align} (d+d_1)+(d_1+d_2)+(d_2+d_3)+\cdots &=d+2d\sum_{n=1}^\infty\left(1\over5\right)^n\\ &=d+2d{1/5\over1-1/5}\\ &=d+2d{1\over5-1}\\ &=d+{2d\over4}\\ &={3\over2}d \end{align}$$

which agrees with what we said earlier: the dog goes fifty percent further than the woman.

Barry Cipra
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  • Could you explain to me the part "Now since the dog runs fifty percent faster than the woman walks, it goes fifty percent further in any given amount of time, meaning $d+\delta={3\over2}(d-\delta)$" ? We have that the speed of the woman is $v$ and the speed of the dog is $\frac{3v}{2}$. How do we get then the above relation? – Mary Star Jan 20 '21 at 19:26
  • @MaryStar, $3/2$ is fifty percent more than $1$. For example, if the woman is walking at four miles per hour, then the dog runs at six miles per hour. So if the woman walks four miles (in an hour), the dog runs six miles (in the same hour). – Barry Cipra Jan 20 '21 at 19:33