Problem 64 - Physics for Scientists and Engineers with Modern Physics

Question

An airplane is heading due south at a speed of \(580 \mathrm{~km} / \mathrm{h}\). If a wind begins blowing from the southwest at a speed of \(90.0 \mathrm{~km} / \mathrm{h}\) (average), calculate (a) the velocity (magnitude and direction) of the plane, relative to the ground, and (b) how far from its intended position it will be after \(11.0 \mathrm{~min}\) if the pilot takes no corrective action. [Hint: First draw a diagram.

Step-by-Step Solution

Answer

The plane's velocity relative to ground is 520.13 km/h, 7.02° west of south. It drifts 11.68 km west in 11 minutes.

Step 1: Define the Vectors and Coordinate System

Let's define the problem in terms of vectors. The airplane's velocity vector can be represented as \( \vec{v}_{plane} = 580 \text{ km/h} \) due south (negative y-direction if north is positive). The wind's velocity is blowing from the southwest, which means it's at a 45-degree angle to both the south and the west directions (i.e., equal components in negative x-direction and positive y-direction).

Step 2: Calculate Wind's Velocity Components

The wind speed is \( 90 \text{ km/h} \). The wind vector components are:\[ v_{wind,x} = 90 \cos(45^\circ) = 90 \frac{\sqrt{2}}{2} \approx 63.64 \text{ km/h} \]\[ v_{wind,y} = 90 \sin(45^\circ) = 90 \frac{\sqrt{2}}{2} \approx 63.64 \text{ km/h} \]Thus, the wind vector is \( \vec{v}_{wind} = (-63.64, 63.64) \text{ km/h} \).

Step 3: Determine the Resultant Velocity Vector

The plane's velocity relative to the ground is the vector sum of the plane's speed and the wind speed:\[ v_{resultant,x} = 0 + (-63.64) = -63.64 \text{ km/h} \]\[ v_{resultant,y} = -580 + 63.64 = -516.36 \text{ km/h} \]Thus, the resultant velocity vector is \( \vec{v}_{resultant} = (-63.64, -516.36) \text{ km/h} \).

Step 4: Calculate Resultant Velocity Magnitude

Use the Pythagorean theorem to find the magnitude of the resultant vector:\[ v_{resultant} = \sqrt{(-63.64)^2 + (-516.36)^2} \approx \sqrt{4050.74 + 266657.58} \approx \sqrt{270708.32} \approx 520.13 \text{ km/h} \]

Step 5: Determine the Direction of the Resultant Velocity

The direction \( \theta \) from the south can be found using the tangent function:\[ \tan\theta = \frac{|-63.64|}{|-516.36|} = \frac{63.64}{516.36} \]\[ \theta \approx \arctan(0.123) \approx 7.02^\circ \] Thus, the plane is heading \( 7.02^\circ \) west of south.

Step 6: Calculate the Displacement in 11 Minutes

Convert time from minutes to hours: \( 11 \text{ min} = \frac{11}{60} \text{ hr} \).Compute the distance traveled in the x and y directions:\[ x_{displacement} = v_{resultant,x} \times \frac{11}{60} = -63.64 \times \frac{11}{60} \approx -11.68 \text{ km} \]\[ y_{displacement} = v_{resultant,y} \times \frac{11}{60} = -516.36 \times \frac{11}{60} \approx -94.03 \text{ km} \]The plane will be approximately 11.68 km west of its intended south path.

Key Concepts

Vector Addition

Vector addition is a fundamental concept in physics, especially when dealing with motion. It involves combining two or more vectors to find a resultant vector that represents the total effect. This is like finding the net result of all the components acting together. In our exercise, we need to add the airplane's velocity vector and the wind's velocity vector.

The airplane's velocity vector points directly south with a magnitude of 580 km/h, and the wind's velocity vector comes from the southwest, creating components along both the north-south and east-west axes. By adding these vectors, we calculate the airplane's velocity relative to the ground.

To handle such additions, break each vector into its components along the coordinate axes. Then, sum the components in the same direction. This way, you end with a new vector representing the combined effect of the original vectors. Vector addition can always be visualized using a parallelogram method or by arranging the vectors head-to-tail.

Velocity Components

Whenever we want to understand how different forces or motions affect an object, splitting the velocity into components is crucial. These components express how much of the velocity is directed along each axis of a coordinate system.

In our scenario, the wind blows from the southwest, meaning that part of its effect will push the airplane sideways, and part will counteract its southward motion. To calculate these effects, we decompose the wind's velocity of 90 km/h into its horizontal (east-west) and vertical (north-south) components.

• Horizontal component (\(v_{wind,x}\)): This was calculated using \(90 \cos(45^\circ)\), resulting in about 63.64 km/h westward.
• Vertical component (\(v_{wind,y}\)): Similarly, \(90 \sin(45^\circ)\), also resulted in approximately 63.64 km/h northward.

These components are crucial for determining how the wind alters the plane's path.

Displacement Calculation

Given the resultant velocity of the plane, it's essential to calculate how far the plane will travel after a certain time. This involves finding the displacement, which needs both the velocity and the time duration.

The plane's resultant velocity components relative to the ground are \(-63.64\) km/h in the x-direction and \(-516.36\) km/h in the y-direction. These values mean the plane is moving west and slightly less south than intended. We calculate displacement over 11 minutes, converting the time into hours since velocity is in km/h.

• Displacement in x-direction: \(-63.64 \times \frac{11}{60} \approx -11.68\) km. This means the plane drifts west from its original path by 11.68 km.
• Displacement in y-direction: \(-516.36 \times \frac{11}{60} \approx -94.03\) km. The plane travels southward by this distance.

Understanding displacement helps pilots make necessary corrections to stay on the intended flight path.

Wind Effect on Motion

Wind significantly impacts the motion of any airborne vehicle. It's crucial to consider its effect during flight navigation. In our problem, the wind was blowing from the southwest, exerting a force that changed the direction and speed of the airplane relative to the ground.

The change induced by wind is not straightforward because it doesn't just add speed or subtract it. Instead, it shifts the entire velocity vector of the plane, altering not only how fast it travels but also in which direction. Pilots often have to adjust for this by changing their heading to counteract the effect of the wind.

• Magnitude and Direction Change: The wind reduced the plane’s southward speed and introduced a westward drift.
• Importance of Calculating Effects: Understanding these influences is crucial for navigation and ensuring the plane reaches its destination.

This makes mastering vector addition and component breakdown vital in aviation.