Question 70302: I don't get how to solve this:
John flies from atlant, GA to San Fransisco, CA. It takes 5.6 hours to travel 2100 miles against the wind. At the same time, debbie flies from San Fransisco to Atlanta. Her plane travels with the same average airspeed but, with the tailwind, her flight takes only 4.8 hours.
A). Write a system of equations that time, air speed, and wind speed to distance for each traveler.
B). Solve the system to find the air speed.
C). Find the wind speed.
If you could show me how to set up these equations, that would be great because i have a few of these and the paper is due tomorrow! thanks,
Rene'
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! John flies from atlanta, GA to San Fransisco, CA. It takes 5.6 hours to travel 2100 miles against the wind. At the same time, debbie flies from San Fransisco to Atlanta. Her plane travels with the same average airspeed but, with the tailwind, her flight takes only 4.8 hours.
A). Write a system of equations that time, air speed, and wind speed to distance for each traveler.
John DATA:
distance = 2100 mi ; time = 5.6 hrs ; rate = d/t = 2100/5.6 = 375 mph
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Debbie DATA:
distance = 2100 mi ; time = 4.8 hrs ; rate = d/t = 2100/4.8 = 437.5 mph
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EQUATIONS:
Let air speed be a and wind speed be w; then:
a+w=437.5 mph
a-w=375 mph
B). Solve the system to find the air speed.
Add the two equation to get:
2a = 812.5
air speed = 406.25 mph
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C). Find the wind speed.
Subtract the two equations to get:
2w=62.5
wind speed = 31.25 mph
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Cheers,
Stan H.
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