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An airlane pilit wishes to fy due west. A wind of 80.0 \(\mathrm{km} / \mathrm{h}\) (about 50 \(\mathrm{mi} / \mathrm{h} )\) is blowing toward the south. (a) If the airspeed of the plane (its speed in still air) is 320.0 \(\mathrm{km} / \mathrm{h}\) (about 200 \(\mathrm{mi} / \mathrm{h} )\) , in which direction should the pilot head? (b) What is the speed of the plane over the ground? TIlustrate with a vector diagram.

Short Answer

Expert verified
(a) The pilot should head 14.5° north of west. (b) The plane's ground speed is 310.4 km/h.

Step by step solution

01

Understand the Problem

The pilot's objective is to fly directly west, despite the fact that there is a wind blowing towards the south. To counteract this, the pilot will need to head in a direction slightly north of west to offset the southward component of the wind.
02

Set Up the Vector Components

We need to establish a coordinate system, where the positive x-axis is east and the positive y-axis is north. Thus, the plane's intended heading is west (negative x), and the wind blows south (negative y).
03

Define Plane's Velocity Components

Let the angle between the plane's heading and the west direction be \( \theta \). The velocity components of the plane can be defined as:- westward component: \( 320.0 \cos \theta \)- northward component: \( 320.0 \sin \theta \) (to counteract the southward wind)
04

Equate Northward Component to Wind

Since the wind speed southward is \(80.0\ \text{km/h}\), the northward component of the plane's velocity must equal this to keep the plane flying directly west. Thus, we set:\[320.0 \sin \theta = 80.0\]
05

Solve for \( \theta \)

Dividing both sides by 320 gives:\[\sin \theta = \frac{80.0}{320.0} = 0.25\]Using the inverse sine function:\[\theta = \sin^{-1}(0.25)\] \[\theta \approx 14.5^\circ\]
06

Calculate Ground Speed

The ground speed of the plane is the westward component of its velocity, which remains unaffected by the wind's north-south impact:\[v_{ground} = 320.0 \cos(14.5^\circ)\]\[v_{ground} \approx 320.0 \times 0.97 = 310.4\ \text{km/h}\]
07

Illustrate with a Vector Diagram

Draw a vector pointing west (left) for the plane's airspeed and a vector pointing south (down) for the wind. The resultant vector should point westward and slightly lower (representing the directed heading), with both the intended airspeed and the wind producing the resulting groundspeed.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Components
When dealing with vectors, like the velocities of a plane and the wind, breaking them down into components is crucial.
This allows us to understand their different directional influences separately. In this context, we use a coordinate system where directions are assigned axes.
  • X-axis: represents the east-west direction. Flying west implies a negative value.
  • Y-axis: represents the north-south direction, helping account for any drift due to winds.

To a pilot wishing to fly due west, the vector components are especially important. They need to counteract the southward drift from the wind.
By splitting the velocity into westward and northward components, we can adjust the plane's heading. It ensures that the actual flight path remains westward.

This process involves calculating trigonometric components: the cosine function helps find the westward component, and the sine function finds the northward one.
Understanding vector components is integral in adjusting initial headings to achieve accurate navigational goals.
Trigonometry in Physics
Trigonometry is a vital tool in physics, used to resolve vector problems like our pilot example.
It provides a way to relate angles and distances in problems with different directions.
  • Sine and Cosine: These trigonometric functions allow us to split vectors into components.
  • Inverse Functions: Used for finding angles when sides are known, as seen with the \(\sin^{-1}\) function applied to determine the angle \(\theta\).

In our exercise, we calculated the angle \(\theta\) that the plane must fly north of west. This angle counteracts the wind's southward force.
The formula used was \(\sin \theta = \frac{80.0}{320.0}\). Solving it provides the specific angle needed.
After determining \(\theta\), trigonometry aids again in finding the ground speed using the cosine function to assess westward movement.

Using trigonometry simplifies complex directional problems, effectively leading to solutions like calculating the shortest path or trajectory correction.
It's an essential part of creating a mathematical model for any situation involving angles and directional changes.
Problem Solving in Physics
Physics often involves solving real-world scenarios using mathematical tools. Such is the case when correcting a flight path.
  • Understand the Objective: Break down what you want to achieve—in this case, flying due west despite a southward wind.
  • Set Up the Problem: Establish variables and known quantities, such as wind speed and desired heading.
  • Use Appropriate Tools: Vectors and trigonometry were key in analyzing and resolving the flight path problem.

Each step involves making informed decisions based on calculated data.
Solving for the angle \(\theta\) and the ground speed required precise trigonometric calculations and component analysis.
Furthermore, using vector diagrams can provide a visual representation. It helps in understanding resultants from vector calculations.

Overall, problem solving in physics encourages a methodical approach.
It blends theory and practical application, using mathematics as a bridge to understand and navigate physical scenarios.

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