Distances and speeds

Question

Alice and Bob started to walk towards each other’s home and then back to theirs, with steady speeds. Alice passed by a bus station at 25 m away from her home, while at the same time Bob was passing by an abandoned old car. Afterwards, they met at 55 m away from Bob’s home and then they met again at 85 m from Alice’s home. What is the distance between the bus station and the abandoned car?

This must be an easy problem but it's been years since I was dealing with maths & physics!

Trying to set up the equations:

If x is the required distance, S the distance between Alice's & Bob's homes and $t_1$ time units passed since they started walking and until they first encountered the bus and the car, also $V_1$ and $V_2$ Alice's and Bob's speed respectively:

$V_1.t_1 + x + V_2.t_1 = S$

$V_1.t_1 = 25$

Then they met after $t_2$ time units:

$V_1.t_2 + V_2.t_2 = S$ and

$V_2.t_2 = 55$

Finally they met again after $t_3$ time units:

$V_1.t_3 + V_2.t_3 = 3S$ (because each one traveled S plus his part of S).

Also $V_2.t_3 = S + 85$

Then what? Apparently I am missing one equation!

Thank you very much!

Answer 1

You have $V_1t_2=S-55 $, $V_2t_2=55$, $V_1t_3=2S-85$, $V_2t_3=S+85$. Hence $$\frac{V_1}{V_2}=\frac{S-55}{55}=\frac{2S-85}{S+85} $$ Thus $$55(2S-85)=(S+85)(S-55)=S^2+30S-85\cdot 55 $$ $$\Rightarrow 110S=S^2+30S\Rightarrow S=80 $$ (assuming $S>0$). In particular $$\frac{V_1}{V_2}=\frac{25}{55} $$ and so the distance between the bus and the car is $$80-25-25\cdot \frac{55}{25}=0 $$

EDIT: This solution is not acceptable (and not the intended solution) because if the distance between Alice's and Bob's house is $80$, and they are only moving between each other's houses, they cannot meet $85$ meters from Alice's house. The equation $$\frac{V_1}{V_2}=\frac{S-55}{55} $$ is correct because the first meeting will necessarily occur before both Alice and Bob have completed their first journey. However, for the second meeting there are other possible cases:

1) It could be that they meet again before Alice has reached Bob's house in her first trip. In this case, the second meeting occurs when in total Alice has travelled $85$ meters and Bob has travelled $S+85$ meters. We obtain $$ \frac{V_1}{V_2}=\frac{S-55}{55}=\frac{85}{S+85}\Rightarrow S^2+30S-2\cdot 55\cdot 85=0$$ a quadratic equation whose positive root is $$ S=5(\sqrt{383}-3)\approx 82.85$$ Since $82.85<85$, this solution is also unacceptable!

2) The remaining case is the one where they meet again before Bob has reached Alice's house in his first trip. Here at the second meeting in total Bob has travelled $S-85$ meters, while Alice has travelled $2S-85$ meters. We get $$\frac{V_1}{V_2}=\frac{S-55}{55}=\frac{2S-85}{S-85} \Rightarrow S^2-250S+2\cdot 55\cdot 85=0$$ a quadratic equation with roots $5(25\pm \sqrt{251})$, of which only
$$S=5(25+\sqrt{251})\approx 204 > 85 $$ is acceptable because $5(25-\sqrt{251})<85$. The distance between the bus and the car is then \begin{align*}x&=S-25-25\cdot\frac{V_2}{V_1}=S-25-25\frac{55}{S-55}=\frac{(S-25)(S-55)-25\cdot 55}{S-55}=\frac{S^2-80S}{S-55}=\\ &=\frac{250S-2\cdot 55 \cdot 85-80S}{S-55}=\frac{170(S-55)}{S-55}=170 \end{align*} This is the intended solution of the problem. Indeed, there are no other possible cases: at the second meeting either Alice or Bob must have completed at least one journey, but none of them can have completed more than one, else there would have been another meeting inbetween.