Question

A Boeing 787 Dreamliner maintains a constant airspeed of 550 miles per hour (mph) headed due north. The jet stream is \(100 \mathrm{mph}\) in the northeasterly direction. (a) Express the velocity \(\mathbf{v}_{\mathrm{a}}\) of the 787 relative to the air and the velocity \(\mathbf{v}_{\mathrm{w}}\) of the jet stream in terms of \(\mathbf{i}\) and \(\mathbf{j}\). (b) Find the velocity of the 787 relative to the ground. (c) Find the actual speed and direction of the 787 relative to the ground.

Short answer

The velocity of the 787 relative to the ground is \( 50\sqrt{2} \mathbf{i} + (550 + 50\sqrt{2}) \mathbf{j} \) or approximately 554.9 mph at 7.2 degrees east of north.

Step-by-step solution

Express the airspeed of the 787 in vector form

Given that the 787 is moving due north at 550 mph, its velocity relative to the air, \( \mathbf{v}_{\mathrm{a}} \), can be expressed as \( \mathbf{v}_{\mathrm{a}} = 0 \mathbf{i} + 550 \mathbf{j} \).

Express the jet stream velocity in vector form

The jet stream is moving 100 mph in the northeasterly direction, which forms a 45-degree angle with the positive x-axis. We resolve this into vector form using trigonometry. \( \mathbf{v}_{\mathrm{w}} = 100 \cos(45^{\circ}) \mathbf{i} + 100 \sin(45^{\circ}) \mathbf{j} \). Simplifying, we get \( \mathbf{v}_{\mathrm{w}} = 100 \times \frac{\sqrt{2}}{2} \mathbf{i} + 100 \times \frac{\sqrt{2}}{2} \mathbf{j} = 50\sqrt{2} \mathbf{i} + 50\sqrt{2} \mathbf{j} \).

Find the velocity of the 787 relative to the ground

The velocity relative to the ground is the sum of the airspeed vector and the jet stream vector. \( \mathbf{v}_{\mathrm{g}} = \mathbf{v}_{\mathrm{a}} + \mathbf{v}_{\mathrm{w}} \). Substituting in the values we get: \[ \mathbf{v}_{\mathrm{g}} = (0 + 50\sqrt{2}) \mathbf{i} + (550 + 50\sqrt{2}) \mathbf{j} \] \ \ Therefore, \( \mathbf{v}_{\mathrm{g}} = 50\sqrt{2} \mathbf{i} + (550 + 50\sqrt{2}) \mathbf{j} \).

Calculate the actual speed of the 787 relative to the ground

The actual speed is the magnitude of the ground velocity vector \( \mathbf{v}_{\mathrm{g}} \). \[ \| \mathbf{v}_{\mathrm{g}} \| = \sqrt{(50\sqrt{2})^2 + (550 + 50\sqrt{2})^2} \] Simplify the calculations: \[ \| \mathbf{v}_{\mathrm{g}} \| = \sqrt{2500 + (550 + 50\sqrt{2})^2} = \sqrt{2500 + 302500 + 5500\sqrt{2}} = \sqrt{305000 + 5500\sqrt{2}} \] Thus, the actual speed is approximately \( 554.9 \textrm{ mph} \).

Determine the direction of the 787 relative to the ground

The direction is given by the angle of the velocity vector relative to the north direction. Compute the angle \( \theta \) using \( \tan \theta \): \[ \tan \theta = \frac{50\sqrt{2}}{550 + 50\sqrt{2}} \] Evaluate: \[ \theta = \arctan\left(\frac{50\sqrt{2}}{550 + 50\sqrt{2}}\right) \approx 7.2^{\circ} \] This angle is east of north.

Key concepts

Vector Addition

Vector addition is a fundamental concept in physics and mathematics. It's about combining two or more vectors to get a resultant vector. Think of vectors as arrows indicating direction and magnitude. For instance, if you were walking north and a gust of wind was pushing you northeast, the combined effect is found using vector addition.

To add vectors, break them down into components. Using the example from our exercise, the plane's velocity and the jet stream are such vectors. The plane's northward speed (550 mph) is one vector, \(\textbf{v}_\text{a} = 0 \textbf{i} + 550 \textbf{j}\), and the jet stream's speed (100 mph at 45° northeast) is another, \(\textbf{v}_\text{w} = 50\text{√2} \textbf{i} + 50\text{√2} \textbf{j}\). To find the resultant velocity relative to the ground, add the components: \(\textbf{v}_\text{g} = \textbf{v}_\text{a} + \textbf{v}_\text{w} = 50\text{√2}\textbf{i} + (550 + 50\text{√2})\textbf{j}\).

Useful tips:
- Break vectors into horizontal (i) and vertical (j) components.
- Always use trigonometry to find components when angles are involved.
- Add all i components together and all j components together to get your resultant vector.

Trigonometry

Trigonometry plays a crucial role in vector calculations, especially when dealing with angles and directions. It allows us to decompose force, velocity, or any vector into its horizontal and vertical components.

The basic trigonometric functions (sine, cosine, and tangent) are used extensively. In the given exercise, the jet stream is 100 mph at a 45° angle. We use cosine for the horizontal component: \(\text{cos}(45°) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\text{adj}}{100}\), which simplifies to \(\textbf{v}_{\text{w},\text{i}} = 100 \times \frac{√2}{2} = 50√2\). Similarly, use sine for the vertical component: \(\text{sin}(45°) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\text{opp}}{100}\), giving \(\textbf{v}_{\text{w},\text{j}} = 100 \times \frac{√2}{2} = 50√2\).

Remember:
- Use \(cos(θ)\) to find the horizontal component of a vector.
- Use \(sin(θ)\) to find the vertical component of a vector.
- \(\text{tan}(θ) = \frac{\text{opposite}}{\text{adjacent}}\).
These help in changing a vector from magnitude and direction form to component form, simplifying vector addition and subtraction.

Velocity

Velocity is a vector quantity that describes both the speed and direction of an object's motion. It differs from speed, which only tells you how fast something is moving. In the exercise, we have the airspeed and jet stream velocities forming vectors to find the ground velocity.

Combine velocities using vector addition to get the resultant velocity relative to the ground. The magnitude of this resultant vector gives the actual speed, calculated by taking the square root of the sum of the squares of its components: \(\text{Speed} = \text√((50√2)^2 + (550 + 50√2)^2)\).

To find the direction of the resultant velocity, use the angle derived from the tangent function: \(\text{tan}(θ) = \frac{50√2}{550 + 50√2}\), and find \(θ\) using arctan: \(θ = \text{arctan}\bigg(\frac{50√2}{550 + 50√2}\bigg) ≈ 7.2°\) east of north.

Key takeaways:
- Velocity includes both speed and direction.
- Resultant velocity combines multiple velocity vectors.
- Use magnitude and trigonometry to find the resultant speed and direction accurately.
Understanding how velocity vectors interact will help you comprehend motion and direction in various physics applications.