Cycling UphillDate: 7/8/96 at 19:33:35 From: Anna Booth Subject: Cycling Problem Dear Dr. Math, My cyclist has posed two more problems. 1. He takes 2 hours to go from A to B and 2 hours 14 minutes to return to A. He travels at xkm/h uphill, 12km/h on the flat, and 15 km/h downhill. A is 23 kilometres from B. If x and y and z are the times he spends uphill, on the flat, and downhill then x + y + z = 2 hours. ?x + 12y + 15z = 23. I don't know how to handle the unknown speed at which he travels uphill. 2. The cyclist goes from A to B regularly and at the same speed. If he goes 1 km/h slower, it takes him 18 minutes longer; if he goes at km/h faster, he gains 15 minutes. Let x be his average speed. (x+1) and (x-1) are his other speeds. If the time he takes is 't' then time varies from t + 15 to t - 18. I have created wonderfully complicated equations which get me nowhere. Julia Booth Date: 7/9/96 at 17:19:16 From: Doctor Anthony Subject: Re: Cycling Problem For the journey from A to B, let d1 = uphill distance, d2 = level distance and d3 = downhill distance. On the return journey, d3 will be uphill and d1 downhill. We can write down the following equations: d1 + d2 + d3 = 23 (1) d1/x + d2/12 + d3/15 = 2 (total time = 2 hours) (2) d1/15 + d2/12 + d3/x = 2+(7/30) (total time = 2 hours 14 minutes) (3) Subtract (2) from (3). d1(1/15 - 1/x) + d3(1/x - 1/15) = 7/30 d1(x-15)/(15x) +d3(15-x)/(15x) = 7/30 (d3 -d1){(15-x)/(15x)} = 7/30 d3 - d1 = (7/30){15x/(15-x)} d3 - d1 = 7/(30-2x) Now from (1) we have d1/12 + d2/12 + d3/12 = 23/12. (4) Subtract (4) from (2) and we have d1(1/x - 1/12) + d3(1/15 - 1/12) = 1/12 d1(1/x - 1/12) - d3(1/60) = 1/12 (5) Subtract (4) from (3) d1(1/15 - 1/12) + d3(1/x - 1/12) = 19/60 -(1/60)d1 + d3(1/x - 1/12) = 19/60 (6) add (4) and (5) d1(1/x - 1/12 - 1/60) + d3(1/x - 1/12 - 1/60) = 2/5 d1(60-6x)/(60x) + d3(60-6x)/(60x) = 2/5 (d1 + d3)(10-x)/(10x) = 2/5 d1 + d3 = (2/5)(10x)/(10-x) d1 + d3 = 4x/(10-x) We note from this that x < 10. If we let x = 9 this gives d1+d3 = 36. Since this is greater than 23 is not possible. Try x = 8. This gives d1 + d3 = 16. We also have d3-d1 = 7x/(30-2x) = 56/14 = 4, so x = 8 is a possible answer. If we try x = 7 we start getting fractional values for d1 and d3, and also small values of d1 and d3. x = 8 is the one to choose, and with this value of x, d1 + d3 = 16, d3 - d1 = 4, so d1 = 6, d3 = 10 and d2 = 7. (2) Let s = distance A to B and u = normal speed in km/hr. Time at normal speed = s/u. Fast time = s/(u+1) = s/u - 1/4 (15 mins = 1/4 hour) (1) Slow time = s/(u-1) = s/u + 3/10 (18 mins = 3/10 hour) (2) We now have two equations with two unknowns, s and u. Equation (1) can be written s/u - s/(u+1) = 1/4. s(u+1-u)/[u(u+1)] = 1/4 s/{u(u+1)} = 1/4 s = (1/4)u(u+1). Equation(2) can be written s/(u-1) - s/u = 3/10. s(u-u+1)/u(u-1) = 3/10 s/{u(u-1)} = 3/10 s = (3/10)u(u-1). Equating the values of s we get: (1/4)u(u+1) = (3/10)u(u-1) divide out the factor u (1/4)(u+1) = (3/10)(u-1) 10(u+1) = 12(u-1) 10u + 10 = 12u - 12 22 = 2u and so u = 11. This is the normal speed he travels. s = (1/4)u(u+1) = (1/4)*11*12 = 33 km. So the distance from A to B is 33 km. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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