Question

A model airplane is flying horizontally due east at \(10 \mathrm{mi} / \mathrm{hr}\) when it encounters a horizontal crosswind blowing south at \(5 \mathrm{mi} / \mathrm{hr}\) and an updraft blowing vertically upward at \(5 \mathrm{mi} / \mathrm{hr}\) 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

Question: Determine the position vector representing the velocity of a model airplane relative to the ground and its speed. The plane is flying horizontally toward the east at 10 mi/hr and encounters a crosswind blowing south at 5 mi/hr and an updraft vertically upward at 5 mi/hr. Solution: The position vector representing the velocity of the plane relative to the ground is ⟨10, -5, 5⟩ mi/hr, and the speed of the plane relative to the ground is 5√6 mi/hr.

Step-by-step solution

Find the position vector of the plane

To find the position vector representing the velocity of the plane relative to the ground, we need to add the velocities in corresponding directions. The plane is flying horizontally toward the east at 10 mi/hr, encountering a crosswind blowing south 5 mi/hr and an updraft vertically upward at 5 mi/hr. We can represent these velocities as vectors: - East: \(\langle 10, 0, 0 \rangle\) mi/hr - South: \(\langle 0, -5, 0 \rangle\) mi/hr (Add a negative sign for the south direction) - Upward: \(\langle 0, 0, 5 \rangle\) mi/hr Now, we add these vectors to find the position vector representing the velocity of the plane relative to the ground: \(\langle 10, 0, 0 \rangle + \langle 0, -5, 0 \rangle + \langle 0, 0, 5 \rangle = \langle 10, -5, 5 \rangle\) mi/hr. So, the position vector representing the velocity of the plane relative to the ground is \(\langle 10, -5, 5 \rangle\) mi/hr.

Find the speed of the plane

Now we will find the speed of the plane relative to the ground. To do this, we can use the Pythagorean theorem in three dimensions for the position vector \(\langle 10, -5, 5 \rangle\): Speed \(= \sqrt{(10)^2 + (-5)^2 + (5)^2} = \sqrt{100 + 25 + 25} = \sqrt{150} = 5\sqrt{6}\) mi/hr. The speed of the plane relative to the ground is \(5\sqrt{6}\) mi/hr.

Key concepts

Vector Addition

Vector addition is a fundamental concept in physics and engineering that allows us to combine multiple vectors into a single vector. When a plane encounters multiple forces or velocities from different directions, we can use vector addition to determine its resultant velocity.

The process involves adding the corresponding components of the vectors. For instance, if we have two vectors, \(\mathbf{a} = \langle a_x, a_y, a_z \rangle\) and \(\mathbf{b} = \langle b_x, b_y, b_z \rangle\), the sum \(\mathbf{a} + \mathbf{b} = \langle a_x + b_x, a_y + b_y, a_z + b_z \rangle\).

In our example with the model airplane, we are combining the eastward, southward, and upward velocities as three separate vectors. Since the south direction is opposite that of north, we add a negative sign to correctly orient the direction in our vector representation. By summing each corresponding component, we arrive at the position vector that describes the plane's overall velocity.

Velocity

Velocity is a vector quantity that represents both the speed and direction of an object's motion. The term 'position vector' often refers to the vector that denotes an object's location relative to a point of origin, but in the context of motion, it's used to represent the velocity vector.

The magnitude of the velocity vector gives the speed, whereas the direction is given by the vector's orientation. In the airplane example, the distinct velocities in different directions need to be combined to find the overall velocity relative to the ground; this is where vector addition plays a crucial role. Once we know the resultant vector, we can determine the plane's speed and direction of travel with respect to the ground.

Pythagorean Theorem in Three Dimensions

While the Pythagorean theorem is most commonly known for its use in two dimensions to find the length of the hypotenuse of a right-angled triangle, it can also be extended to three dimensions. The theorem then helps us calculate the magnitude of a three-dimensional vector.

Considering a three-dimensional vector \(\mathbf{v} = \langle v_x, v_y, v_z \rangle\), the magnitude is the square root of the sum of the squares of its components: \( |\mathbf{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2} \). This is effectively the Pythagorean theorem in three dimensions.

For the airplane's velocity vector \(\langle 10, -5, 5 \rangle\) mi/hr, we calculate the magnitude of the velocity to find the speed relative to the ground. The calculated speed, \(5\sqrt{6}\) mi/hr, reflects the combination of the horizontal and vertical components of the plane's motion. Understanding this three-dimensional application of the Pythagorean theorem is essential for solving problems related to vector magnitudes in three-dimensional space.