Airplane Wind problem












1












$begingroup$


Airplane flying at 400 mph at an angle of 30 deg encounters a wind. The resultant velocity of the airplane is 475.3 mph at an angle of 27.18 deg. What was direction of the wind.



I set this up as tan (27.18) = (Ry/Rx)



But the total Ry contains the y vector of the plane and the y vector of the wind, and since I don't know either wind magnitude nor direction I end up with 2 unknowns in both Ry and Rx.



Then I thought about using law of cosines, but again, come up with one equation, 2 unknowns.



Any thoughts?






algebra-precalculus vectors







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$endgroup$

















    1












    $begingroup$


    Airplane flying at 400 mph at an angle of 30 deg encounters a wind. The resultant velocity of the airplane is 475.3 mph at an angle of 27.18 deg. What was direction of the wind.



    I set this up as tan (27.18) = (Ry/Rx)



    But the total Ry contains the y vector of the plane and the y vector of the wind, and since I don't know either wind magnitude nor direction I end up with 2 unknowns in both Ry and Rx.



    Then I thought about using law of cosines, but again, come up with one equation, 2 unknowns.



    Any thoughts?






    algebra-precalculus vectors







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Airplane flying at 400 mph at an angle of 30 deg encounters a wind. The resultant velocity of the airplane is 475.3 mph at an angle of 27.18 deg. What was direction of the wind.



      I set this up as tan (27.18) = (Ry/Rx)



      But the total Ry contains the y vector of the plane and the y vector of the wind, and since I don't know either wind magnitude nor direction I end up with 2 unknowns in both Ry and Rx.



      Then I thought about using law of cosines, but again, come up with one equation, 2 unknowns.



      Any thoughts?






      algebra-precalculus vectors







      share|cite|improve this question









      $endgroup$




      Airplane flying at 400 mph at an angle of 30 deg encounters a wind. The resultant velocity of the airplane is 475.3 mph at an angle of 27.18 deg. What was direction of the wind.



      I set this up as tan (27.18) = (Ry/Rx)



      But the total Ry contains the y vector of the plane and the y vector of the wind, and since I don't know either wind magnitude nor direction I end up with 2 unknowns in both Ry and Rx.



      Then I thought about using law of cosines, but again, come up with one equation, 2 unknowns.



      Any thoughts?






      algebra-precalculus vectors




      algebra-precalculus vectors






      share|cite|improve this question













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      asked Dec 16 '14 at 22:05









      user163862user163862

      88521017




      88521017






















          2 Answers
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          1












          $begingroup$

          It may be easier to treat the direction of the plane as along one axis, like the $x$ axis:



          $$vec{v_0} = 400 hat{x}.$$



          Then the new velocity is



          $$vec{v} = 475.3 (cos (-2.82^{circ}) hat{x} + sin (-2.82^{circ}) hat{y}) approx 474.72 hat{x} - 23.38 hat{y}.$$



          So the speed of the wind is



          $$sqrt{74.72^2 + (-23.38)^2} approx 78.29 text{ MPH}$$



          and its direction is



          $$arctan(-23.38/74.72) + 30^{circ} approx -17.37^{circ} + 30^{circ} = 12.63^{circ}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Can you tell me how you would get the -2.82 deg?
            $endgroup$
            – user163862
            Dec 16 '14 at 22:28










          • $begingroup$
            Your direction changes by 2.82 degrees south (27.18 - 30).
            $endgroup$
            – John
            Dec 16 '14 at 22:29



















          0












          $begingroup$

          A fairly standard method is to write all the vectors in their $x$ and $y$ components.



          From the aircraft's speed and direction, $400$ mph at an angle of $30$ degrees,
          you get components $A_x$ and $A_y.$
          From the resulting velocity and direction, $475.3$ mph at an angle of $27.18$ degrees,
          you get components $R_x$ and $R_y.$
          This is all from standard formulas for taking $x$ and $y$ components of vectors, for example $A_x = 400 cos(30^circ),$ $A_y = 400 sin(30^circ).$



          Now you know that original velocity + wind should give the resultant velocity,
          that is, $A + W = R.$
          So now you have
          begin{align}
          A_x + W_x &= R_x,\
          A_y + W_y &= R_y.
          end{align}
          Two equations, two unknowns ($W_x$ and $W_y$).



          Finally, take the components $W_x$ and $W_y$ and turn them back into a speed and direction. Now is the time to bring out the tangent function.
          Don't forget the Pythagorean Theorem for the length.




          The Law of Cosines works too. It does not tell you the angle, but it does tell you the length of the unknown side of the vector triangle.
          Once you have the length you can apply the Law of Sines with two known sides and one known angle to get one unknown angle of the triangle,
          and from that you can determine the direction of the vector.



          You do not always need to be able to solve for two things simultaneously when you have two unknowns!
          If you can do one at a time, that's often enough.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

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            active

            oldest

            votes









            1












            $begingroup$

            It may be easier to treat the direction of the plane as along one axis, like the $x$ axis:



            $$vec{v_0} = 400 hat{x}.$$



            Then the new velocity is



            $$vec{v} = 475.3 (cos (-2.82^{circ}) hat{x} + sin (-2.82^{circ}) hat{y}) approx 474.72 hat{x} - 23.38 hat{y}.$$



            So the speed of the wind is



            $$sqrt{74.72^2 + (-23.38)^2} approx 78.29 text{ MPH}$$



            and its direction is



            $$arctan(-23.38/74.72) + 30^{circ} approx -17.37^{circ} + 30^{circ} = 12.63^{circ}.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Can you tell me how you would get the -2.82 deg?
              $endgroup$
              – user163862
              Dec 16 '14 at 22:28










            • $begingroup$
              Your direction changes by 2.82 degrees south (27.18 - 30).
              $endgroup$
              – John
              Dec 16 '14 at 22:29
















            1












            $begingroup$

            It may be easier to treat the direction of the plane as along one axis, like the $x$ axis:



            $$vec{v_0} = 400 hat{x}.$$



            Then the new velocity is



            $$vec{v} = 475.3 (cos (-2.82^{circ}) hat{x} + sin (-2.82^{circ}) hat{y}) approx 474.72 hat{x} - 23.38 hat{y}.$$



            So the speed of the wind is



            $$sqrt{74.72^2 + (-23.38)^2} approx 78.29 text{ MPH}$$



            and its direction is



            $$arctan(-23.38/74.72) + 30^{circ} approx -17.37^{circ} + 30^{circ} = 12.63^{circ}.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Can you tell me how you would get the -2.82 deg?
              $endgroup$
              – user163862
              Dec 16 '14 at 22:28










            • $begingroup$
              Your direction changes by 2.82 degrees south (27.18 - 30).
              $endgroup$
              – John
              Dec 16 '14 at 22:29














            1












            1








            1





            $begingroup$

            It may be easier to treat the direction of the plane as along one axis, like the $x$ axis:



            $$vec{v_0} = 400 hat{x}.$$



            Then the new velocity is



            $$vec{v} = 475.3 (cos (-2.82^{circ}) hat{x} + sin (-2.82^{circ}) hat{y}) approx 474.72 hat{x} - 23.38 hat{y}.$$



            So the speed of the wind is



            $$sqrt{74.72^2 + (-23.38)^2} approx 78.29 text{ MPH}$$



            and its direction is



            $$arctan(-23.38/74.72) + 30^{circ} approx -17.37^{circ} + 30^{circ} = 12.63^{circ}.$$






            share|cite|improve this answer









            $endgroup$



            It may be easier to treat the direction of the plane as along one axis, like the $x$ axis:



            $$vec{v_0} = 400 hat{x}.$$



            Then the new velocity is



            $$vec{v} = 475.3 (cos (-2.82^{circ}) hat{x} + sin (-2.82^{circ}) hat{y}) approx 474.72 hat{x} - 23.38 hat{y}.$$



            So the speed of the wind is



            $$sqrt{74.72^2 + (-23.38)^2} approx 78.29 text{ MPH}$$



            and its direction is



            $$arctan(-23.38/74.72) + 30^{circ} approx -17.37^{circ} + 30^{circ} = 12.63^{circ}.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 16 '14 at 22:26









            JohnJohn

            22.8k32550




            22.8k32550












            • $begingroup$
              Can you tell me how you would get the -2.82 deg?
              $endgroup$
              – user163862
              Dec 16 '14 at 22:28










            • $begingroup$
              Your direction changes by 2.82 degrees south (27.18 - 30).
              $endgroup$
              – John
              Dec 16 '14 at 22:29


















            • $begingroup$
              Can you tell me how you would get the -2.82 deg?
              $endgroup$
              – user163862
              Dec 16 '14 at 22:28










            • $begingroup$
              Your direction changes by 2.82 degrees south (27.18 - 30).
              $endgroup$
              – John
              Dec 16 '14 at 22:29
















            $begingroup$
            Can you tell me how you would get the -2.82 deg?
            $endgroup$
            – user163862
            Dec 16 '14 at 22:28




            $begingroup$
            Can you tell me how you would get the -2.82 deg?
            $endgroup$
            – user163862
            Dec 16 '14 at 22:28












            $begingroup$
            Your direction changes by 2.82 degrees south (27.18 - 30).
            $endgroup$
            – John
            Dec 16 '14 at 22:29




            $begingroup$
            Your direction changes by 2.82 degrees south (27.18 - 30).
            $endgroup$
            – John
            Dec 16 '14 at 22:29











            0












            $begingroup$

            A fairly standard method is to write all the vectors in their $x$ and $y$ components.



            From the aircraft's speed and direction, $400$ mph at an angle of $30$ degrees,
            you get components $A_x$ and $A_y.$
            From the resulting velocity and direction, $475.3$ mph at an angle of $27.18$ degrees,
            you get components $R_x$ and $R_y.$
            This is all from standard formulas for taking $x$ and $y$ components of vectors, for example $A_x = 400 cos(30^circ),$ $A_y = 400 sin(30^circ).$



            Now you know that original velocity + wind should give the resultant velocity,
            that is, $A + W = R.$
            So now you have
            begin{align}
            A_x + W_x &= R_x,\
            A_y + W_y &= R_y.
            end{align}
            Two equations, two unknowns ($W_x$ and $W_y$).



            Finally, take the components $W_x$ and $W_y$ and turn them back into a speed and direction. Now is the time to bring out the tangent function.
            Don't forget the Pythagorean Theorem for the length.




            The Law of Cosines works too. It does not tell you the angle, but it does tell you the length of the unknown side of the vector triangle.
            Once you have the length you can apply the Law of Sines with two known sides and one known angle to get one unknown angle of the triangle,
            and from that you can determine the direction of the vector.



            You do not always need to be able to solve for two things simultaneously when you have two unknowns!
            If you can do one at a time, that's often enough.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              A fairly standard method is to write all the vectors in their $x$ and $y$ components.



              From the aircraft's speed and direction, $400$ mph at an angle of $30$ degrees,
              you get components $A_x$ and $A_y.$
              From the resulting velocity and direction, $475.3$ mph at an angle of $27.18$ degrees,
              you get components $R_x$ and $R_y.$
              This is all from standard formulas for taking $x$ and $y$ components of vectors, for example $A_x = 400 cos(30^circ),$ $A_y = 400 sin(30^circ).$



              Now you know that original velocity + wind should give the resultant velocity,
              that is, $A + W = R.$
              So now you have
              begin{align}
              A_x + W_x &= R_x,\
              A_y + W_y &= R_y.
              end{align}
              Two equations, two unknowns ($W_x$ and $W_y$).



              Finally, take the components $W_x$ and $W_y$ and turn them back into a speed and direction. Now is the time to bring out the tangent function.
              Don't forget the Pythagorean Theorem for the length.




              The Law of Cosines works too. It does not tell you the angle, but it does tell you the length of the unknown side of the vector triangle.
              Once you have the length you can apply the Law of Sines with two known sides and one known angle to get one unknown angle of the triangle,
              and from that you can determine the direction of the vector.



              You do not always need to be able to solve for two things simultaneously when you have two unknowns!
              If you can do one at a time, that's often enough.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                A fairly standard method is to write all the vectors in their $x$ and $y$ components.



                From the aircraft's speed and direction, $400$ mph at an angle of $30$ degrees,
                you get components $A_x$ and $A_y.$
                From the resulting velocity and direction, $475.3$ mph at an angle of $27.18$ degrees,
                you get components $R_x$ and $R_y.$
                This is all from standard formulas for taking $x$ and $y$ components of vectors, for example $A_x = 400 cos(30^circ),$ $A_y = 400 sin(30^circ).$



                Now you know that original velocity + wind should give the resultant velocity,
                that is, $A + W = R.$
                So now you have
                begin{align}
                A_x + W_x &= R_x,\
                A_y + W_y &= R_y.
                end{align}
                Two equations, two unknowns ($W_x$ and $W_y$).



                Finally, take the components $W_x$ and $W_y$ and turn them back into a speed and direction. Now is the time to bring out the tangent function.
                Don't forget the Pythagorean Theorem for the length.




                The Law of Cosines works too. It does not tell you the angle, but it does tell you the length of the unknown side of the vector triangle.
                Once you have the length you can apply the Law of Sines with two known sides and one known angle to get one unknown angle of the triangle,
                and from that you can determine the direction of the vector.



                You do not always need to be able to solve for two things simultaneously when you have two unknowns!
                If you can do one at a time, that's often enough.






                share|cite|improve this answer









                $endgroup$



                A fairly standard method is to write all the vectors in their $x$ and $y$ components.



                From the aircraft's speed and direction, $400$ mph at an angle of $30$ degrees,
                you get components $A_x$ and $A_y.$
                From the resulting velocity and direction, $475.3$ mph at an angle of $27.18$ degrees,
                you get components $R_x$ and $R_y.$
                This is all from standard formulas for taking $x$ and $y$ components of vectors, for example $A_x = 400 cos(30^circ),$ $A_y = 400 sin(30^circ).$



                Now you know that original velocity + wind should give the resultant velocity,
                that is, $A + W = R.$
                So now you have
                begin{align}
                A_x + W_x &= R_x,\
                A_y + W_y &= R_y.
                end{align}
                Two equations, two unknowns ($W_x$ and $W_y$).



                Finally, take the components $W_x$ and $W_y$ and turn them back into a speed and direction. Now is the time to bring out the tangent function.
                Don't forget the Pythagorean Theorem for the length.




                The Law of Cosines works too. It does not tell you the angle, but it does tell you the length of the unknown side of the vector triangle.
                Once you have the length you can apply the Law of Sines with two known sides and one known angle to get one unknown angle of the triangle,
                and from that you can determine the direction of the vector.



                You do not always need to be able to solve for two things simultaneously when you have two unknowns!
                If you can do one at a time, that's often enough.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 25 at 22:57









                David KDavid K

                55.2k344120




                55.2k344120






























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