Question

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?

Answer 1

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

$\displaystyle \vec{v}=\frac{\vec{x}}{t}$

Where $\vec x$ is the displacement. Both the velocity and displacement are vectors. The displacement can be computed from the above relation as

$\displaystyle \vec{x}=\vec{v}.t$

The plane goes at 400 Km/h on a course of 120° for 2 hours. We can compute the components of the velocity as

$\displaystyle \vec{v_1}=<400\ cos\ 120^o,400\ sin\ 120^o>$

$\displaystyle \vec{v_1}=<-200,\ 200\sqrt{3}>\ km/h$

The displacement of the plane in 2 hours is

$\displaystyle \vec{x_1}=\vec{v_1}.t_1=<-200,200\sqrt{3}>.(2)$

$\displaystyle \vec{x_1}=<-400,400\sqrt{3}>km$

Now the plane keeps the same speed but now its course is 210° for 1 hour. The components of the velocity are

$\displaystyle \vec{v_2}=<400\ cos210^o,400\ sin 210^o>$

$\displaystyle \vec{v_2}=<-200\sqrt{3},-200>km/h$

The displacement in 1 hour is

$\displaystyle \vec{x_2}=\vec{v_2}.t_2=<-200\sqrt{3},-200>.(1)$

$\displaystyle \vec{x_2}=<-200\sqrt{3},-200>km$

The total displacement is the vector sum of both

$\displaystyle \vec{x_t}=\vec{x_1}+\vec{x_2}=<-400,400\sqrt{3}>+<-200\sqrt{3},-200>$

$\displaystyle \vec{x_t}=<-400-200\sqrt{3},400\sqrt{3}-200>km$

$\displaystyle \vec{x_t}=<-746.41\ km,492.82\ km>$

The distance from the airport is the module of the displacement:

$\displaystyle |\vec{x_t}|=\sqrt{(-746.41)^2+492.82^2}$

$\displaystyle |\vec{x_t}|=894.43\ km$