Problem 57 - Algebra and Trigonometry

Question

Velocity The speed of an airplane is 300 \(\mathrm{mi} / \mathrm{h}\) relative to the air. The wind is blowing due north with a speed of 30 \(\mathrm{mi} / \mathrm{h}\) . In what direction should the airplane head in order to arrive at a point due west of its location?

Step-by-Step Solution

Answer

The airplane should head approximately 5.74° south of west.

Step 1: Identify the Problem Context

The airplane's speed is 300 miles per hour relative to the air, and the wind is blowing due north at 30 miles per hour. We need to find the direction the airplane should head to go due west.

Step 2: Set Up the Coordinate System

Let's represent the airplane's velocity as a vector. The positive x-axis points east, the positive y-axis points north. The airplane needs to compensate for the northward wind to maintain a westward direction in net movement.

Step 3: Express Velocities in Vector Form

Write the wind velocity vector as \( \vec{w} = (0, 30) \) and the airplane's velocity relative to the air as \( \vec{v}_a = (300 \cos \theta, 300 \sin \theta) \), where \( \theta \) is the heading angle measured from the east direction.

Step 4: Determine Net Velocity

The net velocity \( \vec{v}_{net} \) should be directed west, which implies \( \vec{v}_{net} = (v_x, 0) \) with \( v_x < 0 \). Calculate \( \vec{v}_{net} = \vec{v}_a + \vec{w} = (300 \cos \theta, 300 \sin \theta + 30) \).

Step 5: Solve for Zero Northward Component

Set the northward (y-component) of \( \vec{v}_{net} \) to zero to ensure movement is due west: \[ 300 \sin \theta + 30 = 0 \] Solve this equation for \( \theta \).

Step 6: Solve for \( \theta \)

Rearrange the equation: \[ 300 \sin \theta = -30 \] \[ \sin \theta = -\frac{30}{300} = -\frac{1}{10} \] Find \( \theta \) using the inverse sine: \[ \theta = \sin^{-1}(-\frac{1}{10}) \approx -5.74^{\circ} \] This means the airplane should head approximately \( 5.74^{\circ} \) south of west.

Key Concepts

Vector Analysis

Vector analysis involves breaking down quantities that have both magnitude and direction. For our problem, the airplane's and wind's velocities can be represented as vectors in a coordinate system.
The x-axis is assigned to the east-west direction, and the y-axis represents the north-south direction.
Vectors allow us to visually and mathematically depict movement in two-dimensional space.Let's dive into how we use vectors in this scenario:

• Airplane's velocity vector: Since the airplane's speed is 300 miles per hour, its velocity relative to the air is represented as a vector with components based on its heading angle \(\theta\): \( \vec{v}_a = (300 \cos \theta, 300 \sin \theta) \).
• Wind velocity vector: The wind is blowing due north at 30 miles per hour, contributing only to the y-component: \( \vec{w} = (0, 30) \).

This kind of breakdown helps simplify the problem, allowing for straightforward calculations to determine the resultant path of the airplane by adding the vector components.

Relative Velocity

Relative velocity describes how the velocity of one object appears from the perspective of another.
In our exercise, the airplane's speed relative to the air is different from its speed relative to the ground due to the wind. Relative velocity is crucial in this scenario because:

• By knowing both the velocity of the airplane and the wind, we can calculate the net velocity \(\vec{v}_{net}\) that considers the wind's effect. This is crucial for determining the actual trajectory of the plane over the ground.
• The key challenge is to find a heading that neutralizes the y-component (northward direction) caused by the wind, to ensure the plane travels due west.

This relationship enhances our understanding of how factors such as wind affect an airplane's path, necessitating calculations to adjust its heading for the desired course.

Angle Calculation

Finding the appropriate heading or angle is central to solving navigation problems like this one.
The goal is to determine the angle at which the airplane should head to counterbalance the wind's effect and achieve a straightforward westward path.Here's the step-by-step process to calculate the right heading angle:

• Net y-component: For the airplane to move strictly west, its net north-south component should be zero.
  This means setting the y-component of the net velocity \(\vec{v}_{net}\) to zero:
  \[ 300 \sin \theta + 30 = 0 \]
• Solving for \(\theta\): Rearrange the equation and solve for \(\theta\):
  \[ 300 \sin \theta = -30 \]
  \[ \sin \theta = -\frac{1}{10} \]
  Applying the inverse sine function:
  \[ \theta = \sin^{-1}(-\frac{1}{10}) \approx -5.74^{\circ} \]

This implies the airplane should adjust its heading approximately 5.74 degrees south of west to counteract the wind and maintain a due west path.