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Chapter 3: Problem 70

A small plane is flying directly west with an airspeed of $30.0 \mathrm{m} / \mathrm{s} .\( The plane flies into a region where the wind is blowing at \)10.0 \mathrm{m} / \mathrm{s}\( at an angle of \)30^{\circ}$ to the south of west. (a) If the pilot does not change the heading of the plane, what will be the ground speed of the airplane? (b) What will be the new directional heading, relative to the ground, of the airplane? (tutorial: flight of crow)

Short Answer

Expert verified
Answer: The ground speed of the airplane is approximately 33.3 m/s, and its new directional heading is approximately 9.5° to the south of west.

Step by step solution

01

Identify the Velocity Vectors

We have two velocity vectors in this problem: the airplane's airspeed and the wind speed. Let's denote the airplane's velocity vector as \(\vec{V_A}\) and the wind's velocity vector as \(\vec{V_W}\): - The airplane's velocity vector \(\vec{V_A}\) has a magnitude of 30 m/s and direction directly west (given). - The wind's velocity vector \(\vec{V_W}\) has a magnitude of 10 m/s and direction 30° to the south of west (given).
02

Break Down the Velocity Vectors into Components

Let's break down both velocity vectors into their components in order to work with them more easily: - For the airplane's velocity \(\vec{V_A}\), since it's flying west (horizontally), its horizontal velocity component is \(30 \mathrm{m/s}\), and its vertical velocity component is 0. - For the wind's velocity \(\vec{V_W}\), let's find its horizontal and vertical components by trigonometry. The horizontal component points westwards and the vertical component points southwards. Since the angle is 30°: Horizontal component: \(V_{W_x} = 10\cos 30^\circ = 10\frac{\sqrt{3}}{2} = 5\sqrt{3}\mathrm{m/s}\) Vertical component: \(V_{W_y} = 10\sin 30^\circ = 10\frac{1}{2} = 5\mathrm{m/s}\)
03

Calculate the Ground Velocity Vector Components

The effective velocity is the sum of the airplane's velocity vector and the wind's velocity vector. To find the components of the ground velocity vector, we should add up the horizontal and vertical components: Horizontal component (westwards): \(V_{G_x} = V_{A_x} + V_{W_x} = 30 + 5\sqrt{3}\mathrm{m/s}\) Vertical component (southwards): \(V_{G_y} = V_{A_y} + V_{W_y} = 0 + 5\mathrm{m/s}\)
04

Find the Ground Velocity Vector Magnitude and Direction

To find the ground velocity magnitude, we can use the Pythagorean theorem: \(V_G = \sqrt{{V_{G_x}}^2 + {V_{G_y}}^2} = \sqrt{(30 + 5\sqrt{3})^2 + 5^2} \approx 33.3 \mathrm{m/s}\) To find the ground velocity direction (the airplane's new heading), we can use the inverse tangent function: \(\theta = \tan^{-1}(\frac{V_{G_y}}{V_{G_x}}) = \tan^{-1}(\frac{5}{30 + 5\sqrt{3}}) \approx 9.5^\circ\) south of west.
05

(a) Ground Speed

The ground speed of the airplane is \(V_G \approx 33.3 \mathrm{m/s}\).
06

(b) New Directional Heading

The new directional heading of the airplane, relative to the ground, is approximately \(9.5^\circ\) to the south of west.

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