Question:
An air plane flies 20 km in a direction 60 degrees north of east, then 30 km straight east, then 10 km straight north. find magnitude and direction of resultant displacement.
Answer:
34.33 degree north of east is the direction of resultant displacement and 27.32 km , its magnitude.
Explanation:
Let us consider,
An aero plane in a Northern component at 60 degrees = sin (60)
An aero plane in a Eastern component at 60 degrees = cos (60) Find the x-coordinate at point C,
[tex]x=O A \cos \theta+A B=20\left(\cos 60^{\circ}\right)+30=20(0.5)+30=10+30=40 \mathrm{km}[/tex]
Find the y-coordinate at point C,
[tex]y=O A \sin \theta+B C=20\left(\sin 60^{\circ}\right)+10=20(0.866)+10=17.32+10=27.32 \mathrm{km}[/tex]
Now, displacement, [tex]D=\sqrt{x^{2}+y^{2}}=\sqrt{40^{2}+27.32^{2}}=\sqrt{1600+746.38}=\sqrt{2346.38}=48.439 \mathrm{km}[/tex]
To find direction,
[tex]\tan \varphi=\frac{y}{x}=\frac{27.32}{40}=0.683[/tex]
[tex]\varphi=\tan ^{-1}(0.683)=34.33^{\circ}[/tex]
So, 34.33 degree north of east compared to the starting location.
