Q62P - University Physics with Modern Physics

Question

A plane leaves the airport in Galisteo and flies 170 km at east of north; then it changes direction to fly 230 km at $\mathbf{36}\mathbf{°}$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 crew should fly at a distance of 351km and the plane will land at $11.8°$south of east.

Step 1: Identification of given data

The given data can be listed below as-

• The displacement vector $\stackrel{\rightharpoonup }{A} \mathrm{is} , \stackrel{\rightharpoonup }{A}=170\mathrm{km}$.
• The displacement vector $\stackrel{\rightharpoonup }{B} \mathrm{is}, \stackrel{\rightharpoonup }{B}=230\mathrm{km}$.
• The resultant displacement vector $\stackrel{\rightharpoonup }{R} \mathrm{is}, \stackrel{\rightharpoonup }{R}=\stackrel{\rightharpoonup }{A}+\stackrel{\rightharpoonup }{B}$.

Step 2: 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 and R is the resultant vector.

Step 3: Determine the magnitude and direction in which plane will land

The free body diagram has been provided below-

                   

Analysing the above figure, the x and y components are $\cos \theta$and $\sin \theta$ respectively.

 

The equation of the resultant vector in the x direction can be evaluated by,

$R_x=\left(A_x+B_x\right)$

Here, $A_x \mathrm{and} B_x$ is the resultant vector in the x direction of the vector A and B.

 

Substituting the values in the above equation,

 

$$\begin{gathered} R_x=\left(170 \mathrm{km}\right)\sin 68°+\left(230\mathrm{km}\right)\cos 36° \\ =343.7 \mathrm{km} \end{gathered}$$

 

The equation of the resultant vector in the y direction can be evaluated by,

$R_y=A_y+B_y$

 

Substituting the values in the above equation,

 

$$\begin{gathered} R_y=\left(170 \mathrm{km}\right)\cos 68°-\left(230\mathrm{km}\right)\sin 36° \\ =-71.5\mathrm{km} \end{gathered}$$

 

The equation of the resultant vector is expressed as,

 

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

 

Here, R is the resultant vector

 

Substituting the values in the above equation,

 

$$\begin{gathered} R=\sqrt{{\left(353.7\right)}^2+{\left(-71.5\mathrm{km}\right)}^2} \\ =351\mathrm{km} \end{gathered}$$

 

Thus, the crew should fly at a distance of 351km.

 

The equation of the direction in which plane will land is given as,

 

$\tan \theta =\frac{R_y}{R_x}$

 

Here, $\tan \theta$ is the angle in which the plane will land

 

Substituting the values in the above equation,

 

$$\begin{gathered} \tan \theta =\frac{-71.5\mathrm{km}}{343.7\mathrm{km}} \\ \tan \theta =-0.208 \\ \theta ={\tan }^{-1}\left(-0.208\right) \\ =-11.8° \end{gathered}$$ 

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