Question

question 3 10 pts an airplane undergoes the following displacements: first, it flies in a direction 30° east of north. next, it flies due south. finally, it flies 30° north of west. using analytical methods, determine how far the airplane ends up from its starting point. edit view format table 12pt paragraph b i u a t²

Explanation:

Step1: Define the vectors

Let the magnitudes of the three - displacement vectors be \(d_1\), \(d_2\), and \(d_3\). Assume the first displacement vector \(\vec{A}\) with magnitude \(d_1\) at an angle \(\theta_1 = 30^{\circ}\) east of north, the second displacement vector \(\vec{B}\) with magnitude \(d_2\) in the negative - y direction (\(\theta_2 = 270^{\circ}\)), and the third displacement vector \(\vec{C}\) with magnitude \(d_3\) at an angle \(\theta_3=120^{\circ}\) counter - clockwise from the positive x - axis. In component form, \(\vec{A}=d_1\sin30^{\circ}\hat{i}+d_1\cos30^{\circ}\hat{j}\), \(\vec{B}=-d_2\hat{j}\), \(\vec{C}=-d_3\cos30^{\circ}\hat{i}+d_3\sin30^{\circ}\hat{j}\).

Step2: Sum the x - components

The sum of the x - components of the vectors, \(R_x=d_1\sin30^{\circ}-d_3\cos30^{\circ}\).

Step3: Sum the y - components

The sum of the y - components of the vectors, \(R_y=d_1\cos30^{\circ}-d_2 + d_3\sin30^{\circ}\).
Since the problem does not give the magnitudes of the displacements, assume \(d_1 = d_2=d_3 = d\) (for the sake of finding the net displacement in terms of a single variable).
\(R_x=d\times\frac{1}{2}-d\times\frac{\sqrt{3}}{2}=\frac{d(1 - \sqrt{3})}{2}\)
\(R_y=d\times\frac{\sqrt{3}}{2}-d + d\times\frac{1}{2}=\frac{d(\sqrt{3}+1 - 2)}{2}=\frac{d(\sqrt{3}-1)}{2}\)

Step4: Calculate the magnitude of the resultant vector

The magnitude of the resultant displacement \(R\) is given by \(R=\sqrt{R_x^{2}+R_y^{2}}\).
\[

$$\begin{align*} R&=\sqrt{(\frac{d(1 - \sqrt{3})}{2})^{2}+(\frac{d(\sqrt{3}-1)}{2})^{2}}\\ &=\sqrt{\frac{d^{2}(1 - 2\sqrt{3}+3)}{4}+\frac{d^{2}(3 - 2\sqrt{3}+1)}{4}}\\ &=\sqrt{\frac{d^{2}(4 - 2\sqrt{3}+4 - 2\sqrt{3})}{4}}\\ &=\sqrt{\frac{d^{2}(8 - 4\sqrt{3})}{4}}\\ &=\sqrt{d^{2}(2-\sqrt{3})}\\ &=d\sqrt{2 - \sqrt{3}}\approx0 \end{align*}$$

\]
If we assume \(d = 1\) (since the relative displacements matter), the magnitude of the resultant displacement \(R = 0\).

Answer:

The airplane ends up at the starting point (assuming the magnitudes of the three displacements are equal). If the magnitudes are not equal, the general formula for the magnitude of the resultant displacement is \(R=\sqrt{(d_1\sin30^{\circ}-d_3\cos30^{\circ})^{2}+(d_1\cos30^{\circ}-d_2 + d_3\sin30^{\circ})^{2}}\)