Cycling Motion ProblemDate: 6/26/96 at 19:3:13 From: Anna Booth Subject: Motion Problem Dear Dr. Math, Please can you help me? A cyclist travels from A to B over a high pass, C. His average speed going up the hill is 9 km/h; he goes down the hill at 24km/h. It takes him one and a quarter hours to go from A to B but one and a half hours to returnfrom B to A. What is the distance from A to B? If x is the time he takes to go from A to C and y is the time from C to B, he travels a distance of 9x + 24y and x + y = 5/4 hours. What do I do now? I don't know how to take into account the return trip taking longer. Julia Booth Date: 6/28/96 at 8:52:30 From: Doctor Mike Subject: Re: Motion Problem Hello, You set this up exactly right and you are almost there. It seems you need to assume (reasonable?) that the cyclist's up-hill and down-hill speeds are independent of the direction of travel. The cyclist travels the 24y km from B back up to C in (24y)/9 hours, and the 9x km from C back down to A in (9x)/24 hours. So, (24y)/9 + (9x)/24 equals the return trip time of 3/2 hours. You can do the rest! The distance is 18 km. Hope this is what you need. -Doctor Mike, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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