Chapter 9.4, Problem 79AYU

To determine

To find: the resultant vector represents the path of the plane relative to the ground. And also find the ground speed of the plane and its direction. force.

Answer to Problem 79AYU

  $\begin{array}{l}{V}_a=323.55i+405.51j\\ \left|V_a\right|=518.77km/hr\\ bearing=N38.56E\end{array}$ .

Explanation of Solution

Given:

Air speed of the planeis $500km/hr$

Bearing $N{45}^{\circ }E$

Speed of wind is $60\text{km}/hr$ with the positive x-axis.

Bearing $N{30}^{\circ }W$

Calculation:

As the Air speed of the plane is $500km/hr$ and its Bearing is $N{45}^{\circ }E$ .

Therefore,

Velocity of plane is $500\cos {45}^{\circ }i+500\sin {45}^{\circ }j$ .

  $\Rightarrow V_w=\frac{500}{\sqrt{2}}i+\frac{500}{\sqrt{2}}j$

And the wind speed of the plane is $60km/hr$ and its Bearing is $N{30}^{\circ }E$ .

Velocity of wind is $60\cos {120}^{\circ }i+60\sin {120}^{\circ }j$ .

  $\Rightarrow V_w=-30i+51.96j$

Therefore, resultant vector is equal to the actual velocity of plane.

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

And the ground speed is $\left|V_a\right|=\sqrt{{323.55}^2+{405.51}^2}=518.77km/hr$ .

Now, to find its direction:

  $\begin{array}{l}\tan \theta =\frac{405.51}{323.55}\\ \therefore \theta =51.44\text{with East}\end{array}$ .

Therefore, bearing is $bearing=N38.56E$

Hence, the direction and magnitude of the resultant vector is $V_a=323.55i+405.51j;\left|V_a\right|=518.77km/hr;bearing=N38.56E$ .