/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 71 A small plane is flying directly... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 71

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. In that region, the pilot changes the directional heading to maintain her due west heading. (a) What is the change she makes in the directional heading to compensate for the wind? (b) After the heading change, what is the ground speed of the airplane?

Short Answer

Expert verified
Answer: The pilot must change the directional heading by 9.46掳 to the north to compensate for the wind, and after the heading change, the ground speed of the airplane will be 38.66 m/s.

Step by step solution

01

Represent given vectors

Introduce your vector notation: - The airspeed vector (without wind): \(\vec{A} = 30.0\,m/s\) (due west) - The wind vector: \(\vec{W} = 10.0\,m/s\) at \(30^{\circ}\) south of west
02

Resolve wind vector into components

Decompose the given wind vector into its horizontal (westward) and vertical (southward) components: - Westward component of the wind vector: \(W_x = W \cos(30^{\circ}) = 10\,\mathrm{m/s} \cdot \cos(30^{\circ}) = 8.66\,\mathrm{m/s}\) - Southward component of the wind vector: \(W_y = W \sin(30^{\circ}) = 10\,\mathrm{m/s} \cdot \sin(30^{\circ}) = 5.00\,\mathrm{m/s}\)
03

Calculate the required airspeed vector components

To calculate the required airspeed vector components to have a due west course, we must counteract the southward wind component with a northward component in the airspeed vector: - Westward component of the required airspeed vector: \(A_x = 30\,\mathrm{m/s}\) (given) - Northward component of the required airspeed vector: \(A_y = -W_y = -5.00\,\mathrm{m/s}\)
04

Calculate the new airspeed vector

Calculate the magnitude and direction of the new airspeed vector using its components: - Magnitude of the new airspeed vector: \(A' = \sqrt{A_x^2 + A_y^2} = \sqrt{(30\,\mathrm{m/s})^2 + (-5\,\mathrm{m/s})^2} = 30.4\,\mathrm{m/s}\) - Direction of the new airspeed vector: \(\theta = \arctan\left(\frac{A_y}{A_x}\right) = \arctan\left(\frac{-5\,\mathrm{m/s}}{30\,\mathrm{m/s}}\right) = -9.46^{\circ}\) (north of west) Now we have the answer to part (a) of the question: The pilot must change the directional heading by \(9.46^{\circ}\) to the north to compensate for the wind.
05

Calculate the ground speed

Add the westward components of the required airspeed vector and the wind vector to get the westward component of the ground speed vector: - Westward component of the ground speed vector: \(G_x = A_x + W_x = 30\,\mathrm{m/s} + 8.66\,\mathrm{m/s} = 38.66\,\mathrm{m/s}\) Since the airplane is flying due west, the ground speed is equal to the westward component of the ground speed vector: Ground speed of the airplane: \(G = 38.66\,\mathrm{m/s}\) Now we have the answer to part (b) of the question: After the heading change, the ground speed of the airplane is \(38.66\,\mathrm{m/s}\).

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