Problem 34 - University Physics with Modern Physics

Question

The nose of an ultralight plane is pointed due south, and its airspeed indicator shows 35 m/s. The plane is in a 10-m/s wind blowing toward the southwest relative to the earth. (a) In a vectoraddition diagram, show the relationship of \(\vec{v}_{P/E}\) (the velocity of the plane relative to the earth) to the two given vectors. (b) Let \(x\) be east and y be north, and find the components of \(\vec{v} _{P/E}\). (c) Find the magnitude and direction of \(\vec{v} _{P/E}\).

Step-by-Step Solution

Answer

The plane's velocity relative to the earth is 42.67 m/s at 80.4° south of west.

Step 1: Understanding Velocity Vectors

First identify the given vectors. The velocity of the plane relative to air, \( \vec{v}_{P/A} \), is 35 m/s due south. The wind velocity, \( \vec{v}_{A/E} \), is 10 m/s toward the southwest. We need these vectors to find the velocity of the plane relative to the earth, \( \vec{v}_{P/E} \).

Step 2: Setting up a Vector Addition Diagram

Draw \( \vec{v}_{P/A} \) as a vertical vector pointing due south. Next, draw \( \vec{v}_{A/E} \), which makes a 45° angle to the south-west direction. Use the head-to-tail method to add \( \vec{v}_{P/A} \) and \( \vec{v}_{A/E} \) for the resultant vector \( \vec{v}_{P/E} \).

Step 3: Breaking Down Wind Vector into Components

Decompose \( \vec{v}_{A/E} \) (10 m/s) into its east-west and north-south components. Since it is directed southwest, \( v_{A/E,x} = -10 \cos(45^\circ) = -7.07 \) m/s (west) and \( v_{A/E,y} = -10 \sin(45^\circ) = -7.07 \) m/s (south).

Step 4: Calculating the Components of Velocity Relative to Earth

The plane's velocity in the y-direction is the sum of its own velocity and the southward component of the wind: \( v_{P/E,y} = -35 + (-7.07) = -42.07 \) m/s. The plane's velocity in the x-direction comes entirely from the wind: \( v_{P/E,x} = -7.07 \) m/s.

Step 5: Calculating the Magnitude of the Resultant Velocity

Use the Pythagorean theorem to calculate the magnitude: \(|\vec{v}_{P/E}| = \sqrt{(-7.07)^2 + (-42.07)^2} \approx 42.67\; \text{m/s}.\)

Step 6: Calculating the Direction of the Resultant Velocity

Calculate the angle relative to the east: \( \theta = \tan^{-1}\left(\frac{-42.07}{-7.07}\right) \approx 80.4^\circ \) south of west.

Key Concepts

Vector Addition

In the world of physics, vector addition is a powerful tool to understand how different forces or directions combine. When the nose of a plane points due south with a speed of 35 m/s and the wind is blowing towards the southwest at 10 m/s, this is a classic case of vector addition in action. To solve the problem and find the plane's velocity relative to the earth (\( \vec{v}_{P/E} \)), we create a vector diagram. The technique often used is called the head-to-tail method. Here's how you do it:

• Draw each vector starting with the first vector from a point. In this scenario, the plane’s velocity (\( \vec{v}_{P/A} \)) is a straight line pointing south.
• Next, from the tip (or head) of this vector, draw the second vector, the wind velocity (\( \vec{v}_{A/E} \)), at a 45° angle southwest.
• Finally, the resultant vector (\( \vec{v}_{P/E} \)) is drawn from the start of the first vector to the end of the last vector.

This head-to-tail method visually represents how differing directions and magnitudes of velocity combine to produce a total velocity.

Vector Components

Understanding the components of a vector is crucial when working with vectors that aren't aligned neatly along an axis. Vectors can be broken down into their perpendicular components—usually along the x (east-west) and y (north-south) axes.Let's break down the wind vector \( \vec{v}_{A/E} \), which is 10 m/s toward the southwest:

• Because it makes a 45° angle with both the south and west directions, we use trigonometric functions to find its components:
• The x-component, \( v_{A/E,x} \), is found by \(-10 \cos(45^\circ) = -7.07 \text{ m/s}\).
• The y-component, \( v_{A/E,y} \), is \(-10 \sin(45^\circ) = -7.07 \text{ m/s}\).

Note that both components are negative, indicating the direction: west and south, respectively. Adding these components helps us understand how the overall motion is distributed along these axes, which is vital for session calculations.

Velocity Magnitude

Calculating the magnitude of a resultant velocity vector is a bit like finding the hypotenuse of a right triangle. After determining the x and y components of \( \vec{v}_{P/E} \), the magnitude is computed using the Pythagorean theorem:\[ |\vec{v}_{P/E}| = \sqrt{(-7.07)^2 + (-42.07)^2} \]Performing the calculations:

• First, square each component: \((-7.07)^2\) and \((-42.07)^2\).
• Add these squared values together.
• Finally, take the square root of this sum to find the magnitude, \( |\vec{v}_{P/E}| \approx 42.67\; \text{m/s} \).

The magnitude tells us how fast the plane is actually moving over the ground, irrespective of direction.

Direction of Velocity

Understanding the direction of a velocity vector is critical in physics. Direction tells you where a vector points relative to a reference direction—in this case, relative to the earth’s axes.To find the direction of \( \vec{v}_{P/E} \):

• The angle \( \theta \) is calculated using the inverse tangent function, \( \theta = \tan^{-1}\left(\frac{-42.07}{-7.07}\right) \).
• Solving this gives \( \theta \approx 80.4^\circ \).
• This angle is measured counter-clockwise, so the resultant direction is 80.4° south of west.

This angle indicates that the plane's path over the ground is primarily towards the south but slightly angled westwards as well. Understanding both magnitude and direction is crucial for a full grasp of how an object's motion translates in space.