A spy plane has enough fuel to fly for 5 hours, and its speed in still air is 250 mph. The plane departs with a 40 mph tailwind and returns to the same airport flying into the same wind. How far from the airport can the plane fly under these conditions?
2006-11-08 03:34:59 · 4 answers · asked by shadowcat_97008 1 in Travel ➔ Air Travel
since the wind the the same there and back, we can cancel out the tail wind in our calculations.
250 mph x 5h = 1250 miles in total. So it's 625 miles with a tail wind and 625 back into the wind.
2006-11-08 03:46:58 · answer #1 · answered by borscht 6 · 0⤊ 0⤋
Assume t1 to be time for outgoing trip, and t2 to be time of incoming trip. The plane can fly for 5 hours, so t1 + t2 = 5. Now assume s1 to be the speed at the outgoing trip which should equal 250 + 40 (tailwind) = 290 mph. In same but opposite sense, the speed s2 for incoming trip would be 250 - 40 (headwind) = 210 mph. Let's assume also that the destination airport is at a distance d.
Now, we know that speed is distance cut over time, so the outgoing speed (s1) should equal the distance (d) cut over time (t1), or s1 = d / t1 (formula A). Similarly, s2 = d / t2 (formula B). From formula A we can get d = s1 * t1, and we can substitute that into formula B to get (after some change of places) t1 / t2 = s2 / s1. In other terms, the ratio t1:t2 is s2:s1 or 210:290 or 21:29. So now you have to distribute the five hours at this ratio. The easiest way is to add the two shares 21+29 = 50. Divide the 5 hours into 50 shares, so a share would be 6 minutes (tenth of an hour or 0.1 hours). So t1 would be 21 shares x 6 minutes = 2 hours and 6 minutes. Similarly, t2 = 29 x 6 min = 2 hours and 54 minutes. You see now that t1 + t2 = 5 hours.
Now, if the outgoing trip is 2 hours and 6 minutes (2.1 hours) long flying at a speed of 290 mph, the destination airport would be 290 x 2.1 = 609 miles. You can do it alternatively by the return trip as 210 x 2.9 = 609 miles.
Enjoy!!
2006-11-08 12:07:45 · answer #2 · answered by imdashti 6 · 0⤊ 0⤋
625 miles at a guess................ I would suggest that the tailwind is irrelevant as outward and inward cancel each other out. He can fly for five hours at overall average speed of 250 mph = 1250 total miles, must turn at half way from the airport = 625..............
2006-11-08 11:39:58 · answer #3 · answered by thomasrobinsonantonio 7 · 0⤊ 0⤋
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2006-11-08 11:41:56 · answer #4 · answered by Lil' Gay Monster 7 · 0⤊ 0⤋