Respuesta :
Answer:
Distance from the airport = 894.43 km
Step-by-step explanation:
Displacement and Velocity
The velocity of an object assumed as constant in time can be computed as
[tex]\displaystyle \vec{v}=\frac{\vec{x}}{t}[/tex]
Where [tex]\vec x[/tex] is the displacement. Both the velocity and displacement are vectors. The displacement can be computed from the above relation as
[tex]\displaystyle \vec{x}=\vec{v}.t[/tex]
The plane goes at 400 Km/h on a course of 120° for 2 hours. We can compute the components of the velocity as
[tex]\displaystyle \vec{v_1}=<400\ cos\ 120^o,400\ sin\ 120^o>[/tex]
[tex]\displaystyle \vec{v_1}=<-200,\ 200\sqrt{3}>\ km/h[/tex]
The displacement of the plane in 2 hours is
[tex]\displaystyle \vec{x_1}=\vec{v_1}.t_1=<-200,200\sqrt{3}>.(2)[/tex]
[tex]\displaystyle \vec{x_1}=<-400,400\sqrt{3}>km[/tex]
Now the plane keeps the same speed but now its course is 210° for 1 hour. The components of the velocity are
[tex]\displaystyle \vec{v_2}=<400\ cos210^o,400\ sin 210^o>[/tex]
[tex]\displaystyle \vec{v_2}=<-200\sqrt{3},-200>km/h[/tex]
The displacement in 1 hour is
[tex]\displaystyle \vec{x_2}=\vec{v_2}.t_2=<-200\sqrt{3},-200>.(1)[/tex]
[tex]\displaystyle \vec{x_2}=<-200\sqrt{3},-200>km[/tex]
The total displacement is the vector sum of both
[tex]\displaystyle \vec{x_t}=\vec{x_1}+\vec{x_2}=<-400,400\sqrt{3}>+<-200\sqrt{3},-200>[/tex]
[tex]\displaystyle \vec{x_t}=<-400-200\sqrt{3},400\sqrt{3}-200>km[/tex]
[tex]\displaystyle \vec{x_t}=<-746.41\ km,492.82\ km>[/tex]
The distance from the airport is the module of the displacement:
[tex]\displaystyle |\vec{x_t}|=\sqrt{(-746.41)^2+492.82^2}[/tex]
[tex]\displaystyle |\vec{x_t}|=894.43\ km[/tex]