SOLUTION: An airplane takes 6 hours to travel a distance of 3960 km against the wind. The return trip takes 5 hours with the wind. What is the rate of the plane in still air and what is the
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Question 316321: An airplane takes 6 hours to travel a distance of 3960 km against the wind. The return trip takes 5 hours with the wind. What is the rate of the plane in still air and what is the rate of the wind?
Found 2 solutions by nyc_function, solver91311:
Answer by nyc_function(2741) (Show Source): You can put this solution on YOUR website!
I can see that we have a system of linear equations in two variables hidden in this travel application.
Let p = speed of plane
Let w = speed of wind
Objects traveling against the wind tend to slow down a bit and object traveling with the wind move faster. Agree? Try it on your bike on a windy day.
Against the wind:
time = 6 hours
speed = p - w
distance = 3960 km
================
With the wind:
time = 5 hours
speed = p + w
distance = 3960 km
================
From the tables above, we make the following equations:
6(p - w) = 3960
5(p + w) = 3960
6p - 6w = 3960...Equation A
5p + 5w = 3960...Equation B
Can you take it from here?
TIP: Use the substitution method to find p and w.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Against the wind, the actual speed made good was 3960 divided by 6 equals 660 kilometers per hour. The actual speed made good against the wind is also equal to the speed of the airplane in still air MINUS the speed of the wind.
Similarly, considering the return trip:
Add the two equations:
You should be able to get the wind speed on your own from here.
John
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