Thread: [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?

any help please?

Hi

Let be the speed of the chauffeur
be the speed of Mr Smith
be the normal time of Mr Smith's arrival at the station
be the distance between the station and Mr Smith's home
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 in order to be at at the station
Departure from the station at
Arrival at home at

Special day
Departure of the chauffeur at in order to be at at the station (the chauffeur does not know that Mr Smith is already arrived)
Arrival of Mr Smith at the station at
Arrival of the chauffeur at the meeting point at
Arrival of Mr Smith at the meeting point at
Therefore




Arrival of Mr Smith at home at
This is 10 minutes (or 1/6 hour) earlier than usual
Therefore




Introduced in the first equation gives



Mr Smith walked during minutes

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?

any help please?

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.


Grandad