Problem 76 - Fundamentals of Physics

Question

A light plane attains an airspeed of \(500 \mathrm{~km} / \mathrm{h}\). The pilot sets out for a destination \(800 \mathrm{~km}\) due north but discovers that the plane must be headed \(20.0^{\circ}\) east of due north to fly there directly. The plane arrives in \(2.00 \mathrm{~h}\). What were the (a) magnitude and (b) direction of the wind velocity?

Step-by-Step Solution

Answer

The wind's magnitude is approximately 184.64 km/h, and its direction is about 21.7° south of east.

Step 1: Understand the Problem and Given Variables

We know the airspeed of the plane is \(500 \text{ km/h}\), the destination is \(800 \text{ km}\) due north, the plane is heading \(20.0^\circ\) east of north, and it takes \(2.00 \text{ h}\) to arrive. We need to determine the wind's magnitude and direction.

Step 2: Determine the Plane's Ground Velocity

The ground velocity \( \vec{V_g} \) can be calculated using distance over time: \[ V_g = \frac{800 \text{ km}}{2.00 \text{ h}} = 400 \text{ km/h} \text{ due north.}\]

Step 3: Resolve the Plane's Airspeed into Components

The eastward and northward components of airspeed \( \vec{V_a} \) are given by: \[ V_{a_x} = 500 \cos(20.0^\circ) \quad \text{and} \quad V_{a_y} = 500 \sin(20.0^\circ) \] Calculate these components:\[ V_{a_x} \approx 500 \times 0.9397 = 469.85 \text{ km/h north} \]\[ V_{a_y} \approx 500 \times 0.3420 = 171.00 \text{ km/h east.} \]

Step 4: Determine Wind Velocity Components

The wind velocity \( \vec{V_w} \) has components \( V_{w_x} \) and \( V_{w_y} \) and the relationship is given by: \[ V_{g_x} = V_{a_x} + V_{w_x} \quad \text{and} \quad V_{g_y} = V_{a_y} + V_{w_y} \] Given that \( V_{g_y} = 400 \text{ km/h} \) and \( V_{g_x} = 0 \text{ km/h} \), solve:\[ 400 = 469.85 + V_{w_y} \quad \Rightarrow \quad V_{w_y} = 400 - 469.85 = -69.85 \text{ km/h} \] \[ 0 = 171.00 + V_{w_x} \quad \Rightarrow \quad V_{w_x} = -171.00 \text{ km/h} \]

Step 5: Calculate the Wind's Magnitude and Direction

The magnitude of \( \vec{V_w} \) is found using the Pythagorean theorem:\[ |\vec{V_w}| = \sqrt{(-69.85)^2 + (-171.00)^2} \approx 184.64 \text{ km/h} \]The direction \( \theta \) (south of east) is calculated by:\[ \theta = \tan^{-1}\left( \frac{|V_{w_y}|}{|V_{w_x}|} \right) = \tan^{-1}\left( \frac{69.85}{171.00} \right) \approx 21.7^\circ \]

Key Concepts

Wind Velocity

Wind velocity represents the speed and direction of wind affecting the plane's motion. In aviation, differentiating between airspeed and wind velocity is crucial. While airspeed is the speed of a plane relative to air, wind velocity describes the external wind forces.
Imagine the plane is carrying a constant speed through the air, but the surrounding air (wind) is also moving. This results in the ground velocity, which is the actual speed of the plane over the earth's surface.
Understanding wind velocity helps pilots adjust their course to reach the destination efficiently, knowing how the wind might push them off their path.

Airspeed Components

Breaking down an airspeed vector into components is essential to analyze its direction and magnitude properly.
An airspeed vector can be decomposed into two perpendicular components using trigonometric functions: eastward (horizontal) and northward (vertical) components.

• Eastward Component: Calculated as the product of airspeed and the cosine of the angle from north (here, it's east of north).
• Northward Component: Determined by multiplying the airspeed by the sine of the angle.

Each component illustrates how much of the airspeed is directed towards the east and north, helping in deducing additional effects like wind.

Pythagorean Theorem

Often in physics and navigation, the Pythagorean theorem is utilized to find the resultant magnitude of combined motion vectors.
When assessing resultant velocities, such as wind velocity, we combine its components using this mathematical theorem.
Given the perpendicular components (eastward and northward), the magnitude of the wind velocity is derived as:\[|\vec{V_w}| = \sqrt{(-69.85)^2 + (-171.00)^2}\] This results in a "c"-like hypotenuse representing the wind's overall influence on the plane. It provides a clear picture of how strong and in what general direction the wind is affecting motion.

Trigonometric Functions

Trigonometric functions are vital for breaking down vectors and solving navigation problems. These functions (sine, cosine, tangent) help resolve vectors into components and understand angles.

• Sine (\( \sin \)): Relates to the opposite side of a given angle in a right triangle. Useful for calculating vertical or northward components.
• Cosine (\( \cos \)): Links to the adjacent side, calculating horizontal or eastward components.
• Tangent (\( \tan \)): The ratio of sine over cosine, useful for finding angles based on opposite and adjacent sides.

In problems where airplanes or any objects sway under external forces, these functions depict how forces like wind impact navigation. Thus, mastering these trigonometric functions is essential for understanding vector analysis in real-world contexts.