SOLUTION: A man flies a small airplane a distance of 180 mi. Because he is flying into head wind, the trip takes him 2 hours. On the way back the wind is still blowing at the same speed so t

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Question 1045720: A man flies a small airplane a distance of 180 mi. Because he is flying into head wind, the trip takes him 2 hours. On the way back the wind is still blowing at the same speed so the return trip only takes 1hr and 12 min. What is his speed in the still air? How fast is the wind blowing
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!

               SPEED     TIME       DISTANCE
GOING          r-w        t           d
RETURN         r+w        b           d


Basic Travel RATES rule is RT=D to relate constant rate, time, distance.



Solve the system for r and w. Substitute the given values into the formulas to evaluate the values for r and w.

Answer by ikleyn(52787)   (Show Source): You can put this solution on YOUR website!
.
A man flies a small airplane a distance of 180 mi. Because he is flying into head wind, the trip takes him 2 hours.
On the way back the wind is still blowing at the same speed so the return trip only takes 1 hr and 12 min.
What is his speed in the still air? How fast is the wind blowing
~~~~~~~~~~~~~~~~~~~~~~~

When the man covered 180 mi in 2 hours flying into head wind, the speed of the plane relative to the ground was  = 90 mph. 

This speed is the difference of the plane speed in the still air "u" and the wind's speed "v":

 = 90 = u - v.     (1)

When the man covered the same distance of 180 mi in 1 hour 12 minutes in the return trip flying with the wind, 
the speed of the plane relative to the ground was  = 150 mph. 
Here 1.2 = 1.2 hour = 1 hour and 12 minutes.  (12 minutes =  hour = 0.2 hour).

This speed is the sum of the plane speed in the still air "u" and the wind's speed "v":

 = 150 = u + v.   (2)


So you have this two equations to find u and v:

u - v =  90    (1')   
u + v = 150.   (2')

The simplest way to solve this system is to add the equations (1') and (2').
If you do, you will get

2u = 90 + 150,  or  2u = 240.  Hence,  u =  = 120.

Thus the speed of the plane in the still air is 120 mph.

Having this, you can easily find the speed of the wind v from the equation (1'):

v = u = 90 = 120 - 90 = 30 mph.

Answer.  The speed of the plane is still air is 120 mph.
         The speed of the wind is 30 ph.

Solved.

What I wrote here is the standard way of solving this type of problems and the standard way of explaining the solution.

For more solved problems of this type see the lessons
    - Wind and Current problems
    - More problems on upstream and downstream round trips
    - Wind and Current problems solvable by quadratic equations
    - Unpowered raft floating downstream along a river
    - Selected problems from the archive on the boat floating Upstream and Downstream
    - Selected problems from the archive on a plane flying with and against the wind
in this site.

Please be informed that you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.


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