Answer to Question #246789 in Physics for Bellinson

Question

an airplane is flying in calm air 80 m/s bearing 40 degrees when it suddenly encounters a 20 m/s wind bearing 140. Determine the resultant velocity of the airplane.

Answer 1

$v_{1x}=80\cdot \cos40°=61.28\ (m/s)$

$v_{1y}=80\cdot \sin40°=51.42\ (m/s)$

$v_{2x}=20\cdot \cos140°=-15.32\ (m/s)$

$v_{2y}=20\cdot \sin140°=12.86\ (m/s)$

$v_x=61.28-15.32=45.96 \ (m/s)$

$v_y=51.42+12.86=64.28 \ (m/s)$

$v=\sqrt{45.96^2+64.28^2}=79\ (m/s)$

$\alpha=\tan^{-1}\frac{64.28}{45.96}=54.43°$