Chapter 9, Problem 112RE

To determine

To find: The speed and the direction of the plane relative to the ground.

Answer to Problem 112RE

The speed of the plane is $35.56°$ in the north at an angle of $35.56°$ .

Explanation of Solution

Given:

The speed of the airplane is $500$ km per hour in north direction.

The velocity of the wind is $60$ km per hour in south east direction.

Calculation:

Consider the expression for the velocity of the plane is,

  $\begin{array}{c}{V}_p=500\cos 45°i+500\cos 45°j\\ =\frac{500}{\sqrt{2}}i+\frac{500}{\sqrt{2}}j\end{array}$

Consider the velocity of the wind is,

  $\begin{array}{c}{V}_w=60\cos \left(90+30\right)i+60\cos \left(90+30\right)j\\ =60\cos 120°i+60\sin 120°j\\ =-30i+51.96j\end{array}$

Consider the resultant vector is,

  $\begin{array}{c}{V}_a=V_w+V_p\\ =\frac{500i}{\sqrt{2}}i+\frac{500i}{\sqrt{2}}j-30i+51.96j\\ =323.55i+405.51j\end{array}$

Then, the speed of the ground is,

  $\begin{array}{c}{V}_a=\sqrt{{323.55}^2+{405.51}^2}\\ =518.77\text{km}/\text{hr}\end{array}$

Consider the direction of the flight is,

  $\begin{array}{c}\theta ={\tan }^{-1}\frac{405.51}{323.55}\\ =51.44°\end{array}$

Thus, the direction is in the north as,

  $90°-51.44°=35.56°$