A plane takes off from an airport and flies at a speed of 400km/h on a course of 120° for 2 hours. the plane then changes its course to 210° and continues in this direction for 1 hour. How far is the plane from the airport at the end of this time?

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]

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