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 ```I have a math problem. I've often taken the United to Kona, Hawaii flight and they play a game called halfway to hawaii. The pilots give you the following data: total miles to Kona from SF, average speed of each half of the flight, headwind/tailwind for each half (which you simply add/subtract from speed), exact takeoff and estimated landing time. What is the algebraic formula to solve this problem?? The last time I took the trip, the following facts resulted in a the halfway point at 10:03 AM Hawaii time (I couldn't solve for it even with the facts!): 7:30 AM takeoff (HI time), 5 hrs 15 minutes total flight time, 485 MPH for first half/462 for second, 33 and 45 MPH head wind (so subtract) for 1st/2nd half. Thanks for helping as it's been driving me crazy that I can't figure it out.```
 ```Hi, Thank you for your interesting question! I'd love to fly to Hawaii and have fun answering maths questions at the same time :) Anyways here is the calculation, which should allow you to answer what you're looking for. First you need to calculate the total distance covered. Lets call this distance 'd'. We know that the time needed to travel half of d at 452 miles per hour (485-33) added to the time needed to travel half of d at 417 miles per hour (462-45) totals 5.25 hours Since time = distance / speed we have: d x ( 1/(2 x 452) + 1/(2 x 417)) = 5.25 hours d x ( 1/904 + 1/834) = 5.25 hours d x (834+904)/(834x904) = 5.25 hours d = 5.25 x 834 x 904 / (834+904) = 2277.42 miles Half way is thus: 2277.42/2 = 1138.71 miles Time needed to fly 1138.71 miles at 452 miles per hour: 1138.71 / 252 = 2.519 hours = 2 hours 31 minutes 9 seconds Thus if we add this to the takeoff time of 7:30 we get 10:01. Give or take 30 seconds from each (since the times are not given to the second) and take into account that the speeds are approximate, then you have your answer. I hope this answers your question. If something is unclear of if you need any clarifications, please do not hesitate to ask. Thanks. endo``` Request for Answer Clarification by downtheline-ga on 14 Dec 2003 21:11 PST ```Endo, Hmmm. thanks. I have a couple of points of clarification. Looking at my notes from the info we were given, the pilot also told us that the nautical miles from SFO to Kona were 2097 (not the 2277 that you calculated). He also mentioned that he was proving some "extraneous" (not needed) info. Could his total miles number be a ruse? also, I'm sure I'm just being dense but what is the significance of the 252 in your calculation to get to 2.519 hours? lastly, if time=distance/speed are you simply making the algebraic leap (my math memory is fuzzy) to get to d X speed by "multiplying the inverse" or 1/(2X452)+1/(2x417)? thanks again. tim``` Clarification of Answer by endo-ga on 14 Dec 2003 21:55 PST ```Hi, I think the value given by the pilot is the shortest distance between SFO and Kona rather than the flight path the plane takes and the distance it travels. So indeed I suppose it's a ruse. Secondly it would be too easy otherwise. You would just do: distance/(2 x speed) = 2097/(2x452) = 2.31 hours = 2 hours 19 minutes Which is wrong. Sorry the 252 is a mistake (typo), it's supposed to be 452, the speed during the first half, 485-33 = 452. Sorry for the confusion. Sorry about that last part. I'll rewrite that to make it clearer. So we have time = distance / speed distance travelled = d/2 + d/2 time needed to travel first half: d/(2x452) time needed to travel second half: d/(2x417) time needed in total: d/(2x452) + d/(2x417) = d x ( 1/(2 x 452) + 1/(2 x 417)) (factor out d) Then the rest is the same. I hope this is clearer. If you require any further calculations please do not hesitate to ask. Thanks. endo```