Answer:[tex]\theta =79.47 North\ of\ west[/tex]
Explanation:
Given
Mass of airplane =21500 kg
Force due to jet engines=35100 N
Force from wind is 14900 N at angle of 75 south of west
Resolving forces
[tex]F_x=14900cos75=-3856.403 N[/tex]
[tex]F_y=35100-14900sin75[/tex]
[tex]F_y=35100-14392.29=20707.70 N[/tex]
Therefore acceleration in x direction
[tex]a_x=\frac{F_x}{m}=\frac{-3856.403}{21500}=-0.179 m/s^2[/tex]
[tex]a_y=\frac{F_y}{m}=\frac{20707.70}{21500}=0.963 m/s^2[/tex]
Direction of acceleration
[tex]tan\theta =\frac{a_y}{a_x}=\frac{0.963}{0.179}=-5.38[/tex]
[tex]\theta =79.47\ North\ of\ west[/tex]