Question

A model airplane is flying horizontally due north at \(20 \mathrm{mi} / \mathrm{hr}\) when it encounters a horizontal crosswind blowing east at \(20 \mathrm{mi} / \mathrm{hr}\) and a downdraft blowing vertically downward at \(10 \mathrm{mi} / \mathrm{hr}\). a. Find the position vector that represents the velocity of the plane relative to the ground. b. Find the speed of the plane relative to the ground.

Short answer

Answer: The velocity of the airplane relative to the ground is 20i + 20j - 10k mi/hr, and the speed is 30 mi/hr.

Step-by-step solution

Represent the given velocities as vectors

The plane's velocity is given as 20 mi/hr due north, which can be represented by a vector \(\textbf{v}_p = 20\hat{\textbf{j}}\) mi/hr The crosswind velocity is given as 20 mi/hr to the east, which can be represented by a vector \(\textbf{v}_c = 20\hat{\textbf{i}}\) mi/hr The downdraft velocity is given as 10 mi/hr downward, which can be represented by a vector \(\textbf{v}_d = -10\hat{\textbf{k}}\) mi/hr

Add the vectors to find the velocity vector relative to the ground

The total velocity relative to the ground can be found by adding the three vectors together: \(\textbf{v}_t = \textbf{v}_p + \textbf{v}_c + \textbf{v}_d = 20\hat{\textbf{j}} + 20\hat{\textbf{i}} + (-10\hat{\textbf{k}})\) So the position vector representing the velocity of the plane relative to the ground is \(\textbf{v}_t = 20\hat{\textbf{i}} + 20\hat{\textbf{j}} - 10\hat{\textbf{k}}\) mi/hr #b. Find the speed of the plane relative to the ground#

Calculate the magnitude of the total velocity vector

Now, we will find the speed of the plane by finding the magnitude of the total velocity vector \(\textbf{v}_t\): Speed = \(||\textbf{v}_t|| = \sqrt{(20\hat{\textbf{i}})^2 + (20\hat{\textbf{j}})^2 + (-10\hat{\textbf{k}})^2} = \sqrt{400 + 400 + 100} = \sqrt{900}\) So the speed of the plane relative to the ground is 30 mi/hr.

Key concepts

Velocity Vector

A velocity vector is a crucial concept in physics and mathematics that describes the speed and direction of an object in motion. Think of it like a mathematical arrow pointing from where something started to where it's going, specifying how fast it's moving along the way.

When dealing with velocity vectors, we can use components to break them down into directions. In our example, the model airplane's velocity is influenced by three factors: its northerly flight, a crosswind blowing to the east, and a downdraft pulling it down.

• The plane’s velocity vector due north is represented as \(20\hat{\textbf{j}}\), indicating 20 mi/hr in the north direction.
• The crosswind velocity is \(20\hat{\textbf{i}}\), meaning 20 mi/hr toward the east.
• The downdraft is \(-10\hat{\textbf{k}}\), which shows a downward velocity of 10 mi/hr.

When we sum these vectors, it gives a resultant vector that tells us the plane's actual velocity relative to the ground. This resultant vector is given by:

\(\textbf{v}_t = 20\hat{\textbf{i}} + 20\hat{\textbf{j}} - 10\hat{\textbf{k}}\).
This unified vector shows the combined effect of all the forces acting on the airplane, demonstrating how vector addition works practically.

Magnitude of Vector

Magnitude is a mathematical term used to describe the length or size of a vector. In everyday terms, we can think of it as how fast something moves regardless of its direction. When speaking about velocity, the magnitude of the velocity vector is often referred to as speed.

To find the magnitude of a three-dimensional vector, like the one representing our airplane's velocity, we use the Pythagorean Theorem in space. If our vector is \( \textbf{v}_t = a\hat{\textbf{i}} + b\hat{\textbf{j}} + c\hat{\textbf{k}} \), its magnitude is calculated by the following formula:

\[ ||\textbf{v}_t|| = \sqrt{a^2 + b^2 + c^2} \]

Applying this to our example, we plug in the values from the velocity vector \( \textbf{v}_t = 20\hat{\textbf{i}} + 20\hat{\textbf{j}} - 10\hat{\textbf{k}} \):

• \(a = 20\)
• \(b = 20\)
• \(c = -10\)

The magnitude calculation becomes:

\[ ||\textbf{v}_t|| = \sqrt{20^2 + 20^2 + (-10)^2} = \sqrt{400 + 400 + 100} = \sqrt{900} = 30 \]

This value represents the speed of the airplane relative to the ground as 30 mi/hr.

Three-Dimensional Vectors

Three-dimensional vectors can give a complete picture of motion in space, detailing movement in three perpendicular directions: horizontal, vertical, and depth. Imagine a vector as an adventurous streak flying through space; its direction spans three dimensions allowing it to fully depict a situation like wind effects on a plane.

For vectors in three dimensions, we symbolize the directions with unit vectors: \(\hat{\textbf{i}}\), \(\hat{\textbf{j}}\), and \(\hat{\textbf{k}}\). Each defines movement along one of the three principal axes:

• \(\hat{\textbf{i}}\) for the x-axis (east-west direction)
• \(\hat{\textbf{j}}\) for the y-axis (north-south direction)
• \(\hat{\textbf{k}}\) for the z-axis (up-down direction)

In the context of our problem, a model airplane moves under three forces, which creates a vector that forms an elegant 3D representation of its movement in space.

Understanding three-dimensional vectors makes analyzing real-world scenarios clearer, whether it’s for aircraft navigation, physics problems, or any situation involving spatial considerations. By assigning values to each component of the vector, we can visualize and calculate its overall effect, such as the resulting velocity of the airplane amidst various environmental forces.