How Long Did She Walk?Date: 06/12/99 at 18:35:17 From: ken logan Subject: Problem Solving (Graduate Level) Dear Dr. Math, A woman usually takes the 5:30 p.m. train, arriving at her station at 6:00, and her husband picks her up and drives her home. One day she took the 5:00 p.m. train, arriving at 5:30 p.m. at the station. She began walking home. Her husband, starting out from home to meet her at the usual time, met her on the way and brought her home 10 minutes before the usual time. How long did the woman walk? Date: 06/12/99 at 19:52:27 From: Doctor Anthony Subject: Re: Problem Solving( Graduate Level) Let d = distance from home to the station v = speed of car u = speed she can walk x = distance the woman walked Time for woman to get home from the station home = x/u + (d-x)/v Usual time to get home = d/v Suppose T1 = time car normally arrives at station. On this occasion the woman starts for home at T1 - 1/2 T1 + d/v = time they normally arrive home - T1 - 1/2 + x/u + (d-x)/v = time they arrive home 10 mins early ------------------------- d/v + 1/2 - x/u - (d-x)/v = 1/6 (difference = 1/6 hour) 1/2 - x/u + x/v = 1/6 x(1/u - 1/v) = 1/3 ...............(1) Consider also the time the car meets the woman while she is walking. It would have continued to reach the station at T1. So we have T1 - 1/2 + x/u + x/v = T1 x(1/u + 1/v) = 1/2 ...............(2) now adding equations (1) and (2) we get 2x/u = 1/2 + 1/3 2x/u = 5/6 x/u = 5/12 and since x/u is the time she is walking, we see that this time is 5/12 hour. The woman walks for 25 minutes. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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