Question 214844: Eddie is flying a jet against the wind on a 570 mile trip. Flying with the same wind speed, Eddie could fly a 270 mile trip in the same amount of time. If Eddie's jet flies at 220 miles per hour in still air, what is the speed of the wind?
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! I DON'T THINK YOU HAVE STATED THE PROBLEM CORRECTLY. FLYING WITH THE WIND, EDDIE SHOULD BE ABLE TO TRAVEL A LOT FARTHER IN THE SAME AMOUNT OF TIME THAN HE COULD FLYING AGAINST THE WIND UNLESS THIS PROBLEM IN IMPLYING THAT HE IS FLYING THE 270 MILE TRIP AT THE SPEED OF THE WIND. ANYWAY, HERE'S HOW TO SET UP THE PROBLEM:
Distance(d) equals Rate(r) times Time(t) or d=rt;r=d/t and t=d/r
Let s=speed of the wind
time travelled against the wind=570/(220-s)
time travelled with he wind=270/(220+s)
Now we are told that the above two times are equal, so:
570/(220-s)=270/(220+s) multiply each side by (220-s)(220+s)
570(220+s)=270(220-s) divide each side by 10 (just to reduce the size of the numbers)
57(220+s)=27(220-s) get rid of parens
12540+57s=5940-27s subtract 12540 from and add 27s to each side
12540-12540+57s+27s=5940-12540-27s+27s collect like terms
84s=-6600
s=-78.6 mph----------------WE CAN'T HAVE NEGATIVE WIND SPEED
CHECK YOU PROBLEM
NOW IF THIS PROBLEM IS INDICATING THAT HE FLYS THE 270 MI TRIP AT THE SPEED OF THE WIND, THEN OUR PROBLEM TO SOLVE IS:
570/(220-s)=270/s multiply each term by s(220-s)
570s=270(220-s) divide each side by 1-
57s=27(220-s) or
57s=5940-27s add 27s to each side
57s+27s=5940
84s=5940
s=70.7 mph-------------------wind speed
Hope this helps---ptaylor
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