Chapter 9.4, Problem 80AYU

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 80AYU

  $\begin{array}{l}{V}_a=328.28i-547.89j\\ \left|V_a\right|=638.71km/hr\\ bearing=S30.92E\end{array}$ .

Explanation of Solution

Given:

Air speed of the planeis $600km/hr$

Bearing $S{30}^{\circ }E$

Speed of wind is $40\text{km}/hr$ with the positive x-axis.

Bearing $S{45}^{\circ }W$

Calculation:

As the Air speed of the plane is $600km/hr$ and its Bearing is $S{30}^{\circ }E$ .

Therefore,

Velocity of plane is $600\cos {\left(90-30\right)}^{\circ }i+600\sin {\left(90-30\right)}^{\circ }j$ .

  $\begin{array}{l}\Rightarrow V_p=600\cos {60}^{\circ }i+600\sin {60}^{\circ }j\\ =300i-516.61j\end{array}$

And the wind speed of the plane is $40\text{km}/hr$ and its Bearing is $S{45}^{\circ }W$ .

Velocity of wind is $40\cos {45}^{\circ }i+40\sin {45}^{\circ }j$ .

  $\begin{array}{l}\Rightarrow V_w=\frac{40}{\sqrt{2}}i-\frac{40}{\sqrt{2}}j\\ =28.28i-28.28j\end{array}$

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

  $\begin{array}{l}{V}_a=V_w+V_p\\ =300i-519.61j+28.28i-28.28j\\ =\left(300+28.28\right)i-\left(519.61+28.28\right)j\\ =328.28i-547.89j\end{array}$

And the ground speed is $\left|V_a\right|=\sqrt{{328.28}^2-{547.89}^2}=638.71km/hr$ .

Now, to find its direction:

  $\begin{array}{l}\tan \theta =-\frac{547.89}{328.28}\\ \therefore \theta =-59.07\text{with East}\end{array}$ .

Therefore, bearing is $bearing=S30.92E$

Hence, the direction and magnitude of the resultant vector is $V_a=328.28i-547.89j;\left|V_a\right|=638.71km/hr;bearing=S30.92E$ .