/*! 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 81 An airplane pilot sets a compass... [FREE SOLUTION] | 91Ó°ÊÓ

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An airplane pilot sets a compass course due west and maintains an airspeed of 220 \(\mathrm{km} / \mathrm{h}\) . After flying for 0.500 \(\mathrm{h}\) , she finds herself over a town 120 \(\mathrm{km}\) west and 20 \(\mathrm{km}\) south of her starting point. (a) Find the wind velocity (magnitude and direction). (b) If the wind velocity is 40 \(\mathrm{km} / \mathrm{h}\) due south, in what direction should the pilot set her course to travel due west? Use the same airspeed of 220 \(\mathrm{km} / \mathrm{h}\) .

Short Answer

Expert verified
Wind velocity is 44.72 km/h at 116.57 degrees from north. Head southeast by 10.48 degrees for due west travel.

Step by step solution

01

Calculate the Ground Velocity

The pilot flies for 0.5 hours, with a calculated air distance of 120 km west and 20 km south. The resultant ground velocity is calculated using \( v_{gx} = \frac{\Delta x}{\Delta t} = \frac{120 \, \mathrm{km}}{0.5 \, \mathrm{h}} = 240 \, \mathrm{km/h} \) and \( v_{gy} = \frac{\Delta y}{\Delta t} = \frac{-20 \, \mathrm{km}}{0.5 \, \mathrm{h}} = -40 \, \mathrm{km/h} \).
02

Decompose Air Velocity due West

The known air velocity west composes only the horizontal component, thus, \( v_{ax} = 220 \, \mathrm{km/h} \) and \( v_{ay} = 0 \).
03

Calculate Wind Velocity

Using vector addition, the wind velocity is obtained by subtracting the air velocity vector and ground velocity vector from one another:\[ v_{wind} = (v_{gx} - v_{ax}, v_{gy} - v_{ay}) \]Calculating the components:\[ v_{wx} = 240 \, \mathrm{km/h} - 220 \, \mathrm{km/h} = 20 \, \mathrm{km/h} \] \[ v_{wy} = -40 \, \mathrm{km/h} - 0 \, \mathrm{km/h} = -40 \, \mathrm{km/h} \]Resulting Wind speed is calculated by: \( v_{w} = \sqrt{20^2 + (-40)^2} = \sqrt{400 + 1600} = \sqrt{2000} = 44.72 \, \mathrm{km/h} \).To find direction,\( \theta = \tan^{-1} \left( \frac{-40}{20} \right) = \tan^{-1} (-2) \approx -63.43^\circ \), meaning it blows 63.43 degrees south of west, or 116.57 degrees clockwise from north.
04

Calculate Required Heading for Airspeed due West

Given wind velocity statements (40 km/h south due to wind), to negate vertical movements and achieve straight westward air directionality: \( v_{ay} - v_{wy} =0 \to v_{ay} = 40 \, \mathrm{km/h} \) For airspeed \( 220 \equiv \sqrt{v_{ax}^2 + 40^2} \to v_{ax} = \sqrt{220^2 - 40^2} \equiv \sqrt{48400-1600} \equiv \sqrt{46800} = 216.098 \approx 216.1\, \mathrm{km/h} \)The heading angle is,\( \theta = \tan^{-1} \left( \frac{40}{216.1} \right) \approx \tan^{-1}(0.185) \approx 10.48^\circ \)The pilot should maintain a heading of 10.48 degrees north of west or 349.52 degrees relative to north.

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Key Concepts

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

Wind Velocity Calculation
In the context of physics, understanding wind velocity is crucial for navigation, especially for pilots. Wind velocity is a vector quantity, meaning it includes both magnitude and direction. To calculate wind velocity in our example, we have to consider the difference between the airplane's ground velocity and its airspeed. The ground velocity is determined by both the inherent airspeed of the aircraft and external influences like wind.

We start by knowing that the pilot flew 120 km west and 20 km south in 0.5 hours. These measurements help calculate the ground velocity components: 240 km/h westward and 40 km/h southward. To find the wind velocity, subtract the air velocity vector from the ground velocity. This results in a wind velocity of 20 km/h horizontally east (since less west is essentially east) and 40 km/h south. The magnitude of this vector is calculated using the Pythagorean theorem, which gives us a wind speed of approximately 44.72 km/h. The angle calculated through the arctangent function is about -63.43 degrees, showing it blows diagonally south of west.
Airspeed and Ground Speed
Airspeed refers to the speed of an aircraft relative to the air through which it is flying. Ground speed, however, reflects the aircraft's velocity over the Earth's surface. The difference between these two speeds shows the effect of wind currents on the flight.

In this example, the airplane's airspeed was set at 220 km/h due west, unaffected by wind initially. However, after 0.5 hours, our pilot discovers being displaced southwards due to wind, demonstrating the real effect of wind on the ground speed.
  • Ground velocity measures how far the plane travels over the land in a specified time.
  • Airspeed is crucial for staying aloft and maintaining control, unaffected by terrestrial landmarks.
  • The difference between these velocities helps detect and quantify the wind's interference.
For solving problems like these, you calculate ground speed using the horizontal and vertical displacements over time and identify the necessary adjustments to maintain a specific course.
Vector Addition
Vector addition is a fundamental concept in physics, playing a vital role in calculating net forces like wind velocity. Vectors possess both magnitude and direction and are represented mathematically as components along the axes, often visualized as arrows in spatial configurations.

In this problem, understanding vector addition allows us to figure out the wind's effect on the airplane. The vector of the airplane's movement due to its airspeed is added to the vector of wind velocity. This combined vector gives us the airplane's ground velocity or actual path over the ground.
  • You start by determining individual components of each vector.
  • Combine similar components (e.g., both horizontal components, then both vertical components).
  • Resulting values give you a new vector showing net magnitude and direction.
Using vector addition, you can model and solve various real-world dynamics, predicting actual effects like wind drifts on flight paths.
Trigonometry in Physics
Trigonometry is an invaluable tool in physics, used to solve problems involving directions and magnitudes. It equips us to decompose vectors into components and calculate angles, critical for resolving forces and velocities.

In this exercise, trigonometry helps find the wind direction and also the required heading angle for the pilot when compensating for wind. By applying trigonometric functions like sine, cosine, and tangent, we calculate angles and lengths associated with vectors.
  • Tangent is used to find the angle when you have opposite and adjacent side lengths of a right triangle, as seen while finding wind direction.
  • Pythagorean theorem helps determine the resultant vector magnitude from its components.
  • Knowing these angles aids in navigating correctly despite environmental impacts.
Ultimately, trigonometry aids navigation by ensuring pilots can adjust headings accurately in response to altered conditions while flying, assisting in maintaining intended courses.

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