Math Problem Statement

A pilot is flying from Daytona Beach to Prescott. The two cities are 1874 miles away from one another, and the angle from Daytona Beach to Prescott is W28N. At takeoff, there is a 12 mph wind blowing N37E that blows constantly throughout the entire 4-hour flight. At what angle should the pilot leave Daytona Beach to fly in a straight line and land in Prescott?

Solution

This problem involves vector addition and trigonometry. We need to calculate the correct angle that the pilot should fly to compensate for the wind blowing during the flight.

Step 1: Break down the vectors

  1. The vector of the plane's desired path:
    The desired direction of flight from Daytona Beach to Prescott is W28NW28^\circ N, which means 28 degrees north of west or 36028=332360^\circ - 28^\circ = 332^\circ from the positive x-axis in standard position (measured counterclockwise from east).

  2. The wind vector:
    The wind is blowing at 12 mph in the direction N37EN37^\circ E, which is 37 degrees east of north or 9037=5390^\circ - 37^\circ = 53^\circ from the positive x-axis.

Step 2: Set up the vector components

For both vectors, we will decompose them into their xx (east-west) and yy (north-south) components.

  1. Plane's desired velocity vector:
    The speed of the plane in still air is not given, but we know it needs to counteract the wind to fly in a straight path. Let's assume the plane's airspeed is vpv_p.

    • Vp=(vpcos332,vpsin332)V_p = (v_p \cos 332^\circ, v_p \sin 332^\circ)
    • Vp=(vpcos28,vpsin28)V_p = (v_p \cos 28^\circ, -v_p \sin 28^\circ)
  2. Wind vector:
    The wind speed is 12 mph, and the direction is 53° from the x-axis.

    • Vw=(12cos53,12sin53)V_w = (12 \cos 53^\circ, 12 \sin 53^\circ)

    Calculating the components: Vw=(12×cos53,12×sin53)=(7.21,9.58)V_w = (12 \times \cos 53^\circ, 12 \times \sin 53^\circ) = (7.21, 9.58)

Step 3: Calculate the required angle

The pilot's path plus the wind vector must result in the desired direction. This means the combined velocity vector of the plane and wind must equal the desired velocity vector toward Prescott. Let the angle the pilot should fly be θ\theta.

  • The new velocity vector is the sum of the plane's heading and the wind: (vpcosθ,vpsinθ)+(7.21,9.58)=(vpcos28,vpsin28)(v_p \cos \theta, v_p \sin \theta) + (7.21, 9.58) = (v_p \cos 28^\circ, -v_p \sin 28^\circ)

To solve for θ\theta, we use vector addition and trigonometric identities to balance both the x- and y-components. I'll now calculate the angle numerically.

The pilot should fly at an angle of approximately 122.23122.23^\circ counterclockwise from the east, or W32.23SW32.23^\circ S (32.23 degrees south of west), to compensate for the wind and land in Prescott.

Would you like further details or explanations?

Here are some follow-up questions:

  1. How can vector addition help in other navigational problems involving wind or currents?
  2. What is the significance of breaking vectors into their x- and y-components?
  3. How does trigonometry apply to solving real-world navigation problems?
  4. What would happen if the wind speed changed during the flight?
  5. How does airspeed differ from ground speed in aviation calculations?

Tip: In vector problems, always ensure your angles and directions are consistent with the coordinate system you are using.

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Math Problem Analysis

Mathematical Concepts

Vector Addition
Trigonometry
Angle Calculation

Formulas

Velocity Vector: V = (v cos(θ), v sin(θ))
Wind Vector: W = (w cos(α), w sin(α))
Vector Addition: Resultant Vector = Velocity Vector + Wind Vector

Theorems

Law of Cosines
Law of Sines

Suitable Grade Level

Grades 11-12, College Level