"It is a trick question. Little Johnnie has to be at his grandmother's house 100 miles away in two hours. He runs into traffic and only averages 25 mph for the first 50 miles. How fast does Johnnie need to drive to make it on time? There must be 2,000 versions of this on applied mathematics tests. (They do still teach it don't they?)"
205, then wait just off the end of her driveway until it's exactly 2 hours? Brett
Well, it's sort of a trick question, because it encourages solvers to take an incorrect short cut. The answer is that there is no way then to get there in two hours. The two hours have been used to go 50 miles. There is no time left.
There actually isn't much difference in lap times for "normal" winds, but in higher winds where people still fly, like, say, 15 mph it's noticeable. If the plane flies 55 mi/hr, and the distance is 140ft + 140 ft = 280 ft,
V
w = 15 mph = 15 x 88/60 = 22.00 ft/s,
V
p = 55 mph = 55 x 88/60 = 80.66... ft/s
Ground Speeds: Upwind V
u = 80.67 - 22.00 = 58.67 ft/s; Downwind V
d = 80.67 + 22.00 = 102.67 ft/s
V = d/t => t = d/v: t
u = 140 ft/ 58.67 ft/s = 2.39 s; t
d = 140/102.67 = 1.36 s
t
tot = 2.39 + 1.36 = 3.75 s in the wind.
t without wind = 280/80.67 ft/s = 3.47 s
'gotta run. 'hope I didn't multiply wrong.
Edit: I may multiply right, but as I ran out of the house, it appears that I did not subtract right. So I have corrected the upwind and downwind speeds and the rest accordingly. 'sorry!