Question 985097


Let &nbsp;<B>u</B>&nbsp; be the plane speed in the still air &nbsp;(in miles per hour)&nbsp; and let &nbsp;<B>v</B>&nbsp; be the wind speed &nbsp;(also in &nbsp;{{{mi/h}}}).

Then the speed of the plane against the wind is &nbsp;<B>u - v</B>&nbsp; (relative to the earth), &nbsp;while the speed with the tailwind is &nbsp;<B>u + v</B>.


Thus you have the system of two equation for two unknowns


{{{system(840 = 4*(u - v),
840 = 3.5*(u + v))}}},


or, &nbsp;dividing the first equation by &nbsp;4&nbsp; and the second equation by &nbsp;3.5


{{{system(u - v = 210,
u + v = 240)}}}.


Add the equations in the last system. &nbsp;You will get


2u = 450.


Hence, &nbsp;<B>u</B> = {{{450/2}}} = 225 {{{mi/h}}}. &nbsp;It is the speed of the plane in still air.


Next, &nbsp;substitute this value of &nbsp;<B>u</B>&nbsp; into the second equation, &nbsp;and you will get &nbsp;<B>v</B> = 15 {{{mi/h}}}&nbsp; for the wind speed. 


<B>Answer</B>. &nbsp;The plane speed is &nbsp;225 {{{mi/h}}}&nbsp; in the still air. &nbsp;The wind speed is &nbsp;15 {{{mi/h}}}.


For more wind and current problems see my lessons in this site

&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>.