Problem 106 - Algebra and Trigonometry Real Mathematics, Real People

Question

A commercial jet is flying from Miami to Seattle. The jet's velocity with respect to the air is 580 miles per hour, and its bearing is \(332^{\circ} .\) The wind, at the altitude of the jet, is blowing from the southwest with a velocity of 60 miles per hour. (a) Draw a figure that gives a visual representation of the problem. (b) Write the velocity of the wind as a vector in component form. (c) Write the velocity of the jet relative to the air as a vector in component form. (d) What is the speed of the jet with respect to the ground? (e) What is the true direction of the jet?

Step-by-Step Solution

Answer

The speed of the jet with respect to the ground is calculated by finding the magnitude of the vector sum of the jet’s velocity and the wind’s velocity. The true direction of the jet is calculated by finding the angle of this resultant velocity with respect to East, and converting this to navigational bearing.

Step 1: Draw the scenario

First draw the given scenario. Make sure the velocities and directions are accurately represented. Use arrows for the vectors.

Step 2: Calculate wind vector

The wind is blowing from the Southwest. That means it is blowing to the Northeast. In vector mathematics, it's always about the direction of the movement. So, this direction could be represented in standard form (counter-clockwise from the East) as \[45^{\circ}\]. So, calculate the x (horizontal) and y (vertical) components as follows: \[v_{wx} = v_w * \cos(45^{\circ})\] and \[v_{wy} = v_w * \sin(45^{\circ})\], where \(v_w\) is the velocity of wind. Substitute \(v_w = 60\) mph to find the values.

Step 3: Calculate jet vector

The bearing of the jet is given as \(332^{\circ}\). First, convert this to standard form: \[(360^{\circ} - 332^{\circ}) = 28^{\circ}\] Now, calculate the x (horizontal) and y (vertical) components in the same way as step 2, but using the jet's airspeed: \[v_{jx} = v_j * \cos(28^{\circ})\] and \[v_{jy} = v_j * \sin(28^{\circ})\], where \(v_j\) is the velocity of the jet. Substitute \(v_j = 580\) mph to find the values.

Step 4: Determine ground speed of the jet

Ground speed of the jet is the sum of the velocity vectors of the jet, and the wind. Calculate as follows: \[V_x = v_{wx} + v_{jx}\] and \[V_y = v_{wy} + v_{jy}\].This gives the velocity vector of the jet relative to ground. Ground speed is the magnitude of this vector and it can be determined by applying Pythagoras Theorem: \[\sqrt {V_x^2 + V_y^2}\]

Step 5: Determine the true direction of the jet

The true direction (or the bearing) can be calculated by finding the angle made by resultant velocity with respect to East. Apply the tangent function in reverse (arctan or atan2 on many calculators) to find the angle: \[\arctan (\frac {V_y}{V_x})\]. This gives the angle in standard form. Convert this into navigational bearing (clockwise from North) for the final bearing.

Key Concepts

Vector Components

In the context of vector addition, understanding vector components is crucial. A vector can be split into two components: horizontal (x-component) and vertical (y-component). This decomposition is helpful when dealing with vectors not aligned along an axis.

Consider the wind vector from the exercise. The wind blows from the southwest with a velocity of 60 miles per hour. The direction "southwest" translates to an angle of 45° in the coordinate plane when measured counterclockwise from the positive x-axis (East).

To convert this vector to its components:

• The x-component (horizontal) is calculated using: \(v_{wx} = v_w \cdot \cos(45^{\circ})\)
• The y-component (vertical) is determined by: \(v_{wy} = v_w \cdot \sin(45^{\circ})\)

By plugging in the wind's speed into these equations, it can be expressed in component form. Essentially, this process allows a vector to be easily added or compared to another vector by just summing up their components.

Trigonometric Functions

Trigonometric functions are essential tools for finding the components of a vector. They relate the angles and sides of right triangles to help solve problems involving vectors.

In vector mathematics, when you know the angle and the magnitude (length) of a vector, you can calculate its components using the functions cosine and sine:

• \(\cos(\theta)\) (cosine of the angle): Gives the ratio of the adjacent side (horizontal component) to the hypotenuse (magnitude of the vector).
• \(\sin(\theta)\) (sine of the angle): Provides the ratio of the opposite side (vertical component) to the hypotenuse.

For example, in our exercise, the jet's airspeed is 580 mph with a bearing of 332°. To use trigonometric functions effectively, the bearing is converted to 28° from the North counterclockwise. Subsequently:

• \(v_{jx} = v_j \cdot \cos(28^{\circ})\)
• \(v_{jy} = v_j \cdot \sin(28^{\circ})\)

These calculations transform a directional speed into tangible x and y components, demonstrating just how crucial trigonometry is in vector resolution.

Ground Speed Calculation

When a jet flies, its ground speed is affected by both its velocity through the air and the wind's velocity. Calculating ground speed involves the addition of the vectors for these two velocities. This is done using the components calculated in the previous steps.

To find the ground speed, you add the respective components of the jet's speed and the wind's speed. This is expressed via:

• Horizontal ground speed component: \(V_x = v_{wx} + v_{jx}\)
• Vertical ground speed component: \(V_y = v_{wy} + v_{jy}\)

Once you have these, the ground speed itself is the magnitude of this resultant vector. This is obtained using the Pythagorean theorem:

• \(\sqrt{V_x^2 + V_y^2}\)

This magnitude gives the true speed of the jet over the ground. Its true direction, measured as an angle with respect to the east or north, can also be computed using the inverse tangent function to calculate the angle resulting from \(\tan^{-1}(\frac{V_y}{V_x})\).

Thus, the ground speed calculation takes into account both airspeed and external wind factors, providing a realistic view of how the jet traverses over the ground.