Question 1036908
.
Making a round trip from Fairview to Cartersville, a distance of 20 miles, a pilot faces 30 mph head wind one way 
and 30 mph tail wind on the return trip. The return trip takes 45 minutes less than the outbound journey. 
Find the speed of the plane in still air.
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<pre>
Let "u" be the speed of the plane in still air, in mph.

Then the plane flies at the speed (u+30) mph with the wind, and
                     at the speed (u-30) against the wind.

The plane spends {{{t[1]}}} = {{{20/(u+30)}}} hours flying with the wind, and 
                 {{{t[2]}}} = {{{20/(u-30)}}} hours flying against the wind.

According to the condition, {{{t[2]}}} - {{{t[1]}}} = {{{3/4}}}  (45 minutes = {{{3/4}}} of an hour), which gives an equation

{{{20/(u-30)}}} - {{{20/(u+30)}}} = {{{3/4}}}.

Now solve it for u. First multiply both sides by the common denominator 4*(u-30)*(u+30) and then simplify. You will get

80*(u+30) - 80(u-30)= 3(u+30)*(u-30),

{{{4800}}} = {{{3u^2 - 2700}}},

{{{3u^2}}} = 4800 + 2700 = 7500,

{{{u^2}}} = {{{7500/3}}} = 2500.

u = {{{sqrt(2500)}}} = 50 mph  (only positive root is acceptable).

<U>Answer</U>. The speed of the plane is 50 mph in still air.
</pre>

There are other similar solved problems in this site on a plane flying with and against the wind. &nbsp;See the lessons 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/travel/Wind-and-Current-problems.lesson>Wind and Current problems</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/travel/Wind-and-Current-problems-solvable-by-quadratic-equations.lesson>Wind and Current problems solvable by quadratic equations</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/travel/Selected-problems-from-the-archive-on-a-plane-flying-with-and-against-the-wind.lesson>Selected problems from the archive on a plane flying with and against the wind</A>