1. [SOLVED] Word Geometry Problem

Mr. Smith takes the train daily from his work in the city back to the suburb where he lives and is met at the station by his chauffeur, who drives
him home. One day Mr. Smith finishes his work earlier than usual and arrives in his suburb one hour earlier than usual. He then walks toward his home from the station, and meets his chauffeur (who was not aware of Mr. Smith's early leaving time) on the way. The chauffeur stops and takes Mr. Smith home; this brings Mr. Smith home 10 minutes earlier than usual. Disregarding time for stopping and picking up, and assuming that the chauffeur normally arrives at the station at the same moment as Mr. Smith, for how long (in minutes) did Mr. Smith walk?

2. Hi

Let $\displaystyle v_c$ be the speed of the chauffeur
$\displaystyle v_s$ be the speed of Mr Smith
$\displaystyle t_0$ be the normal time of Mr Smith's arrival at the station
$\displaystyle d$ be the distance between the station and Mr Smith's home
$\displaystyle d_1$ be the distance from Mr Smith's home to the point where Mr smith was picked up by his chauffeur

Normal day
Departure of the chauffeur at $\displaystyle t_0 - \frac{d}{v_c}$ in order to be at $\displaystyle t_0$ at the station
Departure from the station at $\displaystyle t_0$
Arrival at home at $\displaystyle t_0 + \frac{d}{v_c}$

Special day
Departure of the chauffeur at $\displaystyle t_0 - \frac{d}{v_c}$ in order to be at $\displaystyle t_0$ at the station (the chauffeur does not know that Mr Smith is already arrived)
Arrival of Mr Smith at the station at $\displaystyle t_0 - 1$
Arrival of the chauffeur at the meeting point at $\displaystyle t_0 - \frac{d}{v_c} + \frac{d_1}{v_c}$
Arrival of Mr Smith at the meeting point at $\displaystyle t_0 - 1 + \frac{d-d_1}{v_s}$
Therefore
$\displaystyle t_0 - \frac{d}{v_c} + \frac{d_1}{v_c} = t_0 - 1 + \frac{d-d_1}{v_s}$

$\displaystyle \frac{d_1-d}{v_c} = \frac{d-d_1}{v_s} - 1$

Arrival of Mr Smith at home at $\displaystyle t_0 - \frac{d}{v_c} + \frac{2d_1}{v_c}$
This is 10 minutes (or 1/6 hour) earlier than usual
Therefore
$\displaystyle t_0 - \frac{d}{v_c} + \frac{2d_1}{v_c} = t_0 + \frac{d}{v_c} - \frac{1}{6}$

$\displaystyle d-d_1 = \frac{v_c}{12}$

Introduced in the first equation gives

$\displaystyle \frac{v_c}{v_s} = 11$

Mr Smith walked during $\displaystyle \frac{d-d_1}{v_s} = \frac{v_c}{12\:v_s} = \frac{11}{12} = 55$ minutes

3. Word problem

Hello amma0913
Originally Posted by amma0913
Mr. Smith takes the train daily from his work in the city back to the suburb where he lives and is met at the station by his chauffeur, who drives
him home. One day Mr. Smith finishes his work earlier than usual and arrives in his suburb one hour earlier than usual. He then walks toward his home from the station, and meets his chauffeur (who was not aware of Mr. Smith's early leaving time) on the way. The chauffeur stops and takes Mr. Smith home; this brings Mr. Smith home 10 minutes earlier than usual. Disregarding time for stopping and picking up, and assuming that the chauffeur normally arrives at the station at the same moment as Mr. Smith, for how long (in minutes) did Mr. Smith walk?

See the attached diagram.

The diagram shows the displacement-time graph, where the displacement s is measured from home, O. S is the position of the station. OF is 60 minutes.

On the day in question Mr Smith’s walk is represented by SE, and his car ride by ED.

The car’s normal displacement-time graph is AC, then CB. Note that AF = FB since the time taken for both parts of the journey are equal.

On the early day, the car’s displacement-time graph is AE – ED. So ED is parallel to CB; and DB = 10 minutes.

Through F and H, draw lines FG and HJ parallel to DE, meeting AC at G and J. Then all the triangles AHJ, AFG, ADE and ABC are similar. And triangles AHE and AFC are similar.

So AF:HF = AC:EC = AB: DB. But AB = 2AF. So DB = 2HF. So HF = 5 minutes. So he walks for 60 – 5 = 55 minutes.