Q62 P - University Physics with Modern Physics

Question

A plane leaves the airport in Galisteo and flies 170 km at $68°$ east of north; then it changes direction to fly 230 km at 36$°$ south of east, after which it makes an immediate emergency landing in a pasture. When the airport sends out a rescue crew, in which direction and how far should this crew fly to go directly to this plane?

Step-by-Step Solution

Answer

The plane should head to south-east direction.

The plane should head to south-east direction .

The displacement vector $\overrightarrow{A}$ is $\overrightarrow{A}$ =170km.

The displacement vector $\overrightarrow{B}$ is $\overrightarrow{B}$ = 230km.

The resultant displacement vector $\overrightarrow{R}$ is $\overrightarrow{R}$ = $\overrightarrow{A}+\overrightarrow{B}$.

Concept of vector addition

The addition of two or more vector into a vector sum is called vector addition.

The resultant vector is given as,

$R=\sqrt{A^2+B^2}$

Here, A and B are displacement vector.

Determine the direction in which plane will land

The resultant vector can be evaluated by,

 $R_x=\left(A_x+B_x\right)$ 

 $R_x=\left(170km\right)\cos 68°-\left(230km\right)\sin 36°$

$R_x=343.7km$

$R= \sqrt{R_x^2+R_y^2}$

$R= \sqrt{{\left(343.7km\right)}^2+{\left(-71.5km\right)}^2}$

$R=351km$

The direction in which plane will land,

$\tan \theta =\left|\frac{R_x}{R_y}\right|$

$\tan \theta =\frac{71.5km}{343.7km}$

$\tan \theta = 0.208$

$\theta ={\tan }^{-1}\left(0.208\right)$

$\theta =11.8°$

Thus, the plane will land at $11.8°$ south-east.