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Chapter 3: Problem 38

An airplane pilot wishes to fly 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? Illustrate with a vector diagram.

Short Answer

Expert verified
The pilot should head 14.5掳 north of west; ground speed is 310.4 km/h.

Step by step solution

01

Understand the Problem

The pilot wants to fly due west against a southward wind. The plane's airspeed is 320.0 km/h, and the wind speed is 80.0 km/h. We need to determine the necessary direction for the plane to fly to counteract the wind's effect and determine the resulting ground speed.
02

Set Up the Vector Diagram

Visualize the situation as a right triangle where the plane's actual path is the hypotenuse. The westward component of the plane airspeed is the opposite side, and the wind to the south is the adjacent side. The vectors are: Airspeed vector (320 km/h, needing to have a west component), and Wind vector (80 km/h to the south).
03

Determine the Aircraft Heading

We use vector addition to find out the correct heading. The west component of the plane's velocity should be 320.0 km/h to negate wind's effect. Let's call the direction of airspeed \( \theta \) where: \( \cos \theta = \frac{v_{west}}{v_{plane}} \). Therefore \( v_{west} = 320 \cdot \cos \theta \) and \( v_{south} = 320 \cdot \sin \theta = 80 \).
04

Solve for \( \theta \)

Using the previous relation: \( 320 \cdot \sin \theta = 80 \). Thus, \( \sin \theta = \frac{80}{320} = 0.25\). Solving for \( \theta \), we get \( \theta = \arcsin(0.25) \approx 14.5^\circ \). Thus, the pilot should head 14.5掳 north of west.
05

Determine the Plane's Ground Speed

The ground speed is the magnitude of the actual velocity vector moving due west. Therefore, \( v_{ground} = 320 \cdot \cos \theta \). Substituting \( \cos 14.5^\circ \approx 0.97 \), we get \( v_{ground} = 320 \cdot 0.97 \approx 310.4 \ \mathrm{km/h} \).
06

Create a Vector Diagram

Draw a vector diagram with a westward vector representing 310.4 km/h ground speed, a northward 80 km/h component for the airspeed, and a southern wind vector of 80 km/h. Their vector addition illustrates the resultant heading and ground speed.

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Key Concepts

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

Airspeed Calculation
Understanding an aircraft's airspeed is all about comprehending how fast a plane travels through still air. In our given problem, the airspeed is given as 320 km/h. This number signifies that the plane would move at this speed in the absence of any wind. Airspeed is crucial for pilots to account for, as it establishes the baseline speed before considering external forces like wind.

To determine the correct flight path against varying wind conditions, pilots must calculate airspeed accurately. Here's why this is essential:
  • Navigation Precision: It helps in counteracting wind effects and maintaining a steady course.
  • Fuel Efficiency: Knowing accurate airspeed supports optimal fuel consumption efficiency.
  • Safety: Ensures the plane is within safe operational speeds under fluctuating weather conditions.
Thus, knowing that the airspeed is 320 km/h allows the pilot to determine necessary adjustments when the wind is blowing.
Ground Speed Determination
Ground speed is essentially how fast the airplane moves relative to the ground, which differs from the airspeed when wind is a factor. In the given problem, the wind blows south at 80 km/h, affecting the airplane's actual movement westward. To determine ground speed, vector components need to be analyzed.

By using vector mathematics, we can illustrate this:
  • Use the airspeed vector heading and wind vector to form a right-angled triangle.
  • Mathematically, vector addition will provide the resultant ground speed vector.
  • Since the airplane needs to fly directly west, solve for the westward velocity component.
With the airspeed set at a specific angle north of west to counteract the southward wind, the pilot can now calculate the relative ground speed as approximately 310.4 km/h. This value is fundamental for flight time predictions and route planning.
Pilot Navigation Problem
The pilot navigation problem often involves making crucial adjustments to counteract environmental variables, such as wind. In this exercise, the pilot wishes to fly west while a southern wind affects the planned route. By approaching this problem with vector addition, we achieve more precise navigation.

The problem-solving process involves several steps:
  • Forming a clear vector diagram to visualize the airplane's intended path and the actual affected path due to wind.
  • Utilizing trigonometric principles to determine the correct heading angle, in this case, approximately 14.5掳 north of west.
  • Adjusting the airplane's heading compensates for the wind, ensuring a true westward trajectory.
This method allows pilots to adjust their course effectively, maintaining the desired route despite environmental challenges. Problem-solving skills in navigation are critical, impacting flight efficiency, safety, and accuracy.

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Most popular questions from this chapter

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