An airplane has an airspeed of 500 kilometers per hour bearing N45°E. The wind velocity is 60 kilometers per hour in the direction N30°W. Find the resultant vector representing the path of the plane relative to the ground?

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Answer 1 · 2015-11-09T21:26:53

Use unit vectors to solve the addition of two vectors.

Airplane : $500[cos(45)i+sin(45)j]=500[sqrt2/2i+sqrt2/2j]$

Wind : $60[cos(120)i+sin(120)j]=60[-1/2i+sqrt3/2j]$

Next, sum the two vectors:

Sum $=323.5534i+405.5149j$

Angle of resultant vector: $Tan^-1(405.5149/323.5534)~~51.414^o$
or $~~N38.6^oE$

Magnitude of resultant vector: $sqrt(323.5534^2+405.5149^2)~~518.7766$ km/hr

Here is a sketch of the plane vector $(AB)$ and the wind vector $(AC)$ . The resultant vector is $AD$

hope that helped!

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