Problem 57
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
Verified Answer
The airplane should head with an angle \( \theta = \sin^{-1}\left(\frac{1}{10}\right) \) south of west.
1Step 1: Understand the Scenario
The airplane has a speed of 300 mi/h relative to the air. The wind is blowing north at 30 mi/h. We need to find the heading of the airplane such that its resultant velocity points due west.
2Step 2: Define the Coordinate System
Let's define the positive x-axis as eastward and the positive y-axis as northward. Any westward direction will correspond to a negative x-direction, while the north and south directions correspond to movement along the y-axis.
3Step 3: Set the Vector Equations
Let the angle the plane should head at be \( \theta \), measured from the west. The velocity of the plane relative to the air is broken into two components: \( 300 \cos{\theta} \) (westward) and \( 300 \sin{\theta} \) (northward). The wind affects only the northward component.
4Step 4: Resultant Velocity in North-South Direction
Since we want the net north-south velocity to be zero (westward direction only), set the northward component of the airplane's velocity against the wind: \[ 300 \sin{\theta} = 30 \] solving for \( \theta \), gives \( \sin{\theta} = \frac{1}{10} \).
5Step 5: Determine \( \theta \) for the Heading
Calculate \( \theta \) using the inverse sine function: \( \theta = \sin^{-1}\left(\frac{1}{10}\right) \). This angle corresponds to the airplane's heading direction adjusted for the wind to result in a trajectory that is due west.
Key Concepts
Resultant VelocityTrigonometric FunctionsRelative Motion
Resultant Velocity
When dealing with vector addition, understanding resultant velocity is crucial. Resultant velocity refers to the overall speed and direction that results from adding two or more velocities together. In this context, the velocities involved are the airplane's speed relative to the air and the wind velocity. The goal is to find a final velocity that leads the plane directly west.
This involves combining the velocity of the plane (300 mi/h) with the effect of the northward blowing wind (30 mi/h). By adjusting the angle at which the plane is heading, we can ensure that the northward component of the plane's velocity is counteracted by the wind. The resultant velocity should have no northward component, leaving only the westward movement intact.
The resultant velocity is influenced by the direction in which the plane heads. Each component of the velocity is crucial in forming the resultant, which ideally points due west after accounting for wind influence.
This involves combining the velocity of the plane (300 mi/h) with the effect of the northward blowing wind (30 mi/h). By adjusting the angle at which the plane is heading, we can ensure that the northward component of the plane's velocity is counteracted by the wind. The resultant velocity should have no northward component, leaving only the westward movement intact.
The resultant velocity is influenced by the direction in which the plane heads. Each component of the velocity is crucial in forming the resultant, which ideally points due west after accounting for wind influence.
Trigonometric Functions
Trigonometry plays a vital role in determining the components of the airplane's velocity. It helps us break down the plane's movement into two perpendicular components: one westward (x-direction) and one northward (y-direction). By using trigonometric functions, primarily sine and cosine, these calculations become straightforward.
The cosine function helps determine the westward component of the plane's velocity as it heads at an angle \( \theta \). Thus, \( 300 \cos(\theta) \) represents the westward portion. On the other hand, the sine function calculates the northward part, expressed as \( 300 \sin(\theta) \). These formulas derive from the basic understanding of a right-angled triangle, where the hypotenuse is the airplane's speed, and the other sides are its velocity components.
The cosine function helps determine the westward component of the plane's velocity as it heads at an angle \( \theta \). Thus, \( 300 \cos(\theta) \) represents the westward portion. On the other hand, the sine function calculates the northward part, expressed as \( 300 \sin(\theta) \). These formulas derive from the basic understanding of a right-angled triangle, where the hypotenuse is the airplane's speed, and the other sides are its velocity components.
- Cosine: Determines the component along the horizontal axis.
- Sine: Determines the component along the vertical axis.
Relative Motion
Relative motion is a concept that explains how an object's velocity changes when other moving frames of reference are considered. In this exercise, the relative motion involves considering how the airplane's velocity is perceived when both its speed and the wind velocity align differently.
Despite the wind blowing north, the airplane aims to fly due west. This means we must assess the motion relative to both the air and the ground. By setting the northward components of the velocities against each other, the airplane can maintain a path unaffected by the wind heading directly west.
Despite the wind blowing north, the airplane aims to fly due west. This means we must assess the motion relative to both the air and the ground. By setting the northward components of the velocities against each other, the airplane can maintain a path unaffected by the wind heading directly west.
- Airplane Velocity: Speed of the airplane relative to the air trajectory.
- Wind Velocity: External influence affecting the plane's path.
Other exercises in this chapter
Problem 54
Components of a Velocity A jet is flying in a direction N \(20^{\circ} \mathrm{E}\) with a speed of \(500 \mathrm{mi} / \mathrm{h}\). Find the north and east co
View solution Problem 55
Velocity A river flows due south at 3 mi/h. A swimmer attempting to cross the river heads due east swimming at \(2 \mathrm{mi} / \mathrm{h}\) relative to the wa
View solution Problem 58
Velocity A migrating salmon heads in the direction N \(45^{\circ} \mathrm{E},\) swimming at \(5 \mathrm{mi} / \mathrm{h}\) relative to the water. The prevailing
View solution Problem 59
True Velocity of a Jet A pilot heads his jet due east. The jet has a speed of \(425 \mathrm{mi} / \mathrm{h}\) relative to the air. The wind is blowing due nort
View solution