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You are piloting a small airplane in which you want to reach a destination that is \(750 \mathrm{km}\) due north of your starting location. Once you are airborne, you find that (due to a strong but steady wind) to maintain a northerly course you must point the nose of the plane at an angle that is \(22^{\circ}\) west of true north. From previous flights on this route in the absence of wind, you know that it takes you 3.14 h to make the journey. With the wind blowing, you find that it takes \(4.32 \mathrm{h}\). A fellow pilot calls you to ask about the wind velocity (magnitude and direction). What is your report?

Short Answer

Expert verified
The wind velocity is approximately 101.67 km/h at an angle of 61° west of north.

Step by step solution

01

Calculate the speed of the airplane in still air

First, determine the speed of the airplane without wind. The distance is given by 750 km, and the time taken without wind is 3.14 hours. The speed of the airplane in still air (\text{v}_{\text{plane}}) can be calculated as: \[ \text{v}_{\text{plane}} = \frac{750 \text{ km}}{3.14 \text{ h}} \ \text{v}_{\text{plane}} = 238.22 \text{ km/h} \ \text{(rounded to two decimal places)} \]
02

Calculate the ground speed of the airplane with wind

Next, calculate the ground speed of the airplane when the wind is blowing. The distance is still 750 km, but the time taken with wind is 4.32 hours. The ground speed (\text{v}_{\text{ground}}) can be calculated as:\[ \text{v}_{\text{ground}} = \frac{750 \text{ km}}{4.32 \text{ h}} \ \text{v}_{\text{ground}} = 173.61 \text{ km/h} \]
03

Determine the components of the airplane's velocity with wind

Now, decompose the airplane's velocity into components. Given the airplane must point 22° west of true north to maintain a northerly course, the components are:\[ \text{v}_{\text{plane, north}} = 238.22 \text{ km/h} \times \text{cos}(22^\text{°}) \ \text{v}_{\text{plane, west}} = 238.22 \text{ km/h} \times \text{sin}(22^\text{°}) \ \]
04

Calculate the north and west components

Calculate the north and west components of the airplane's velocity:\[ \text{v}_{\text{plane, north}} = 238.22 \text{ km/h} \times 0.9272 = 220.97 \text{ km/h} \ \text{v}_{\text{plane, west}} = 238.22 \text{ km/h} \times 0.3746 = 89.15 \text{ km/h} \ \]
05

Determine the wind's effect

The airplane's ground speed northward is given by the ground speed in the direction of travel. We can find this by subtracting the north component of the airplane's speed with respect to air from the ground speed:\[ \text{v}_{\text{wind, north}} = \text{v}_{\text{ground}} - \text{v}_{\text{plane, north}} \ \text{v}_{\text{wind, north}} = 173.61 \text{ km/h} - 220.97 \text{ km/h} = -47.36 \text{ km/h} \ \]
06

Calculate the westward wind component

The westward component of the wind's velocity is equal and opposite to the westward component of the airplane's velocity because the plane is stationary (i.e., no movement westward) relative to the ground in the westward direction.\[ \text{v}_{\text{wind, west}} = - \text{v}_{\text{plane, west}} = -89.15 \text{ km/h} \]
07

Calculate the wind's magnitude and direction

Using the Pythagorean theorem, calculate the wind's magnitude:\[ \text{v}_{\text{wind}} = \text{sqrt}(\text{v}_{\text{wind, north}}^2 + \text{v}_{\text{wind, west}}^2) \ \text{v}_{\text{wind}} = \text{sqrt}((-47.36 \text{ km/h})^2 + (-89.15 \text{ km/h})^2) \ \text{v}_{\text{wind}} = 101.67 \text{ km/h} \ \ \text{The wind direction can be found using:} \ \theta_{\text{wind}} = \text{tan}^{-1}\bigg(\frac{\text{v}_{\text{wind, west}}}{\text{v}_{\text{wind, north}}}\bigg) \ \theta_{\text{wind}} = \text{tan}^{-1}\bigg(\frac{89.15 \text{ km/h}}{47.36 \text{ km/h}}\bigg) = 61^\text{°} \ \text{Thus, the wind is blowing } 61^\text{°} \text{ west of north.} \]

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Key Concepts

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

Vector Decomposition
When solving problems in physics, especially in air navigation, it's essential to break down vectors into their respective components. This method, known as vector decomposition, allows us to better understand and manage the different forces and directions involved. In our problem, we decompose the velocity of the airplane into north and west components because the plane is navigating in a windy environment.
For instance, if the plane needs to head 22 degrees west of true north, we need to calculate how much of its velocity points north and how much points west. This helps in understanding how the wind affects the plane's movement and is vital for accurately determining wind velocity.
Velocity Components
Decomposing the velocity of an airplane into its north and west components allows us to see how these directions interact. In the given problem, the plane has a known velocity in still air. We use trigonometric functions, notably sine and cosine, to find these components.
The formulas we use here are:
  • Northward velocity component: \(v_{\text{plane, north}} = v_{\text{plane}} \times \text{cos}(22^\text{°})\)
  • Westward velocity component: \(v_{\text{plane, west}} = v_{\text{plane}} \times \text{sin}(22^\text{°})\)

By using these equations, we calculate how much of the plane's speed aims north and west. This breakdown provides the precise data needed to figure out the wind's impact.
Pythagorean Theorem
Understanding the Pythagorean Theorem is crucial when dealing with wind velocities. In our context, it helps to find the magnitude of the wind's velocity when we know its north and west components. The Pythagorean Theorem states that in a right-angled triangle, the square of the hypotenuse is the sum of the squares of the other two sides.
Mathematically, it is represented as:
\(
v_{\text{wind}} = \text{sqrt}(v_{\text{wind, north}}^2 + v_{\text{wind, west}}^2)
\) .
Using this theorem, after determining the north and west components of the wind's velocity, we calculate the overall wind speed. This provides us with the wind's magnitude, essential for accurate air navigation.
Trigonometric Functions in Physics
Trigonometric functions play a fundamental role in solving physics problems involving angles and directions. In the exercise, we used these functions to find the components of the airplane's velocity. Utilizing sine and cosine, we can accurately split vectors into their respective parts.
Here's an overview of their roles:
  • **Cosine function**: Used for finding the adjacent side of a right-angled triangle, which, in this case, is the north component of the velocity.
  • **Sine function**: Used for finding the opposite side, which is the west component of the velocity.

We start with the airplane's known velocity in still air and multiply it by the cosine of 22 degrees to find the north component and by the sine of 22 degrees to get the west component. Trigonometry thus enables precise and practical solutions to real-world navigation problems.

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