Calculating Distance for Woman & Dog Walking Home

[mathmari]

Hey! :giggle:

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 $(d_n)_{n\geq 1}$.The time that woman needs is equal to $t=\frac{s}{v}$.

At the same time the distance that the dog makes is $s(d)=\frac{3v}{2}\cdot \frac{s}{v}=\frac{3s}{2}$, right?

But how can we get a formula for $(d_n)_{n\geq 1}$ ?

:unsure:

 

[I like Serena]

Hey mathmari!

Let's start with $d_1$.
Let $x$ be the distance of the woman to her home and let $s$ be the distance of the dog to home.
Initially the woman is at $x_0=d$ and I assume the dog is at distance $s_0=0$.
Given the speeds we have $x=d-vt$ and $s=\frac 32v t$ until they meet at $x=s=d_1$ at $t=t_1$.
Can we find $t_1$ and $d_1$?

 

[mathmari]

Let's start with $d_1$.
Let $x$ be the distance of the woman to her home and let $s$ be the distance of the dog to home.
Initially the woman is at $x_0=d$ and I assume the dog is at distance $s_0=0$.
Given the speeds we have $x=d-vt$ and $s=\frac 32v t$ until they meet at $x=s=d_1$ at $t=t_1$.
Can we find $t_1$ and $d_1$?

We get that $$x=s \Rightarrow d_1-vt=\frac{3}{2}vt\Rightarrow d_1=\frac{5}{2}vt_1$$ and substituting this into $x=d-vt$ we get the time, right?

 

[I like Serena]

I think we're mixing up $d$ and $d_1$.
The first is the initial distance, while the second is the distance at which the woman meets her dog for the first time.

 

[mathmari]

I think we're mixing up $d$ and $d_1$.
The first is the initial distance, while the second is the distance at which the woman meets her dog for the first time.

Having that $x=s$ we get $d-vt=\frac{3}{2}vt \Rightarrow \frac{5}{2}vt=d \Rightarrow t=\frac{2d}{5v}$, which is $t_1$, or not?

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

Is that correct so far? :unsure:

 

[mathmari]

After the dog and the woman met, the dog goes back to the house and then again to the direction of the woman.
So at time $t_1$ the position of the dog is $s_1=\frac{3d}{5}$. So we get $s=\frac{3d}{5}-\frac{3v}{2}t$. When $s=0$ the dog is at the house, so at $t=\frac{2d}{5v}$. Then the dog goes to the other direction, back to the woman so $s= \frac{3}{2}vt$ with $t>\frac{2d}{5v}$.
At time $t_1$ the position of the woman is also $x_1=\frac{3d}{5}$. So we get $x=\frac{3d}{5}-vt$, or not?
So to get the next time they meet we have to set $x=s=d_2$, or not? Then we get $$\frac{3d}{5}-vt=\frac{3}{2}vt \Rightarrow \frac{3d}{5}=\frac{5}{2}vt \Rightarrow t=\frac{6d}{25v}$$ Then $$d_2=x(t_2)=\frac{3d}{5}-v\cdot \frac{6d}{25v}=\frac{9d}{25}$$

So the dog and the woman meet at positions $d_n =\left (\frac{3}{5}\right )^nd$.

Is that correct so far? :unsure: