Question 1120244
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Let  S  be the distance between their homes, in meters.

Let  {{{V[1]}}}  be Alice's rate, and  {{{V[2]}}} be Bob's rate.

Let  {{{t[2]}}}  be the time after their start to their first meeting.

Let  {{{t[3]}}}  be the time after their start to their second meeting.


     Notice I follow your  (your ?)  designations in your  (your ?)  set up in

          <A HREF=https://math.stackexchange.com/questions/2859310/distances-and-speeds>https://math.stackexchange.com/questions/2859310/distances-and-speeds</A>

          https://math.stackexchange.com/questions/2859310/distances-and-speeds


Then you have 

    {{{V[1]*t[2]}}} = S - 55;     {{{V[2]*t[2]}}} = 55;           (1)

    {{{V[1]*t[3]}}} = 2S - 85;     {{{V[2]*t[3]}}} = S + 85.      (2)


From (1) you have  {{{V[1]/V[2]}}} = {{{(S-55)/55}}};               (3)

From (2) you have  {{{V[1]/V[2]}}} = {{{(2S-85)/(S+85)}}},              (4)

which gives you an equation for S


    {{{(S-55)/55}}} = {{{(2S-85)/(S+85)}}}.


Simplify and solve for S:

    (S-55)*(S+85) = (2S-85)*55

    S^2 -55*S + 85*S - 55*85 = 110*S - 55*85

    S^2 - 55*S + 85*S - 110*S = 0

    S^2 - 80*S = 0

    S*(S-80) = 0  ====>  The only positive root is   S = 80.


Thus we found the distance between the homes: it is 80 meters.


Having this known,  you van calculate the ratio of their rates from (3)

    {{{V[1]/V[2]}}} = {{{(80-55)/55}}} = {{{25/55}}} = {{{5/11}}}   

and hence calculate the distance "x" from the Bob's home to the abandoned car from the proportion

    {{{25/x}}} = {{{5/11}}}  ======>  x = {{{(25*11)/5}}} = 55 m.


Then the distance under the question is  80 - 25 - 55 meters = 0 meters.
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