High School

An airplane flies in the direction of 119 degrees at a speed of 718 km/h. The wind blows from the southwest at 111 km/h at a direction of 66 degrees. What is the resulting speed in km/h with respect to the ground?

Answer :

The resulting speed of the airplane with respect to the ground is approximately 777.0 km/h and it is flying in a direction of approximately 67.1 degrees south of west.

Step 1: Understanding the Directions:

  • The airplane flies at 119 degrees, which means it is heading slightly south of east.
  • The wind blows from the southwest at 111 km/h at 66 degrees, which means the wind is actually blowing towards the northeast at that angle.

Step 2: Breaking Down the Velocities into Components:

We will split both the airplane's and the wind's velocities into their horizontal (x) and vertical (y) components.

  1. Airplane's Velocity

    • Speed = 718 km/h
    • Angle = 119 degrees

    [tex]V_{Ax} = 718 \cdot \cos(119^{\circ})[/tex]
    [tex]V_{Ay} = 718 \cdot \sin(119^{\circ})[/tex]

  2. Wind Velocity

    • Speed = 111 km/h
    • Angle = 66 degrees (from the north towards the east) means that we need to adjust for the direction. The components are:

    [tex]V_{Wx} = 111 \cdot \cos(66^{\circ})[/tex]
    [tex]V_{Wy} = 111 \cdot \sin(66^{\circ})[/tex]

Step 3: Calculate the Components:

Now we compute the individual components.

  1. For the airplane:

    • [tex]V_{Ax} = 718 \cdot \cos(119^{\circ}) \approx -339.6 \text{ km/h}[/tex]
    • [tex]V_{Ay} = 718 \cdot \sin(119^{\circ}) \approx 616.1 \text{ km/h}[/tex]

  2. For the wind:

    • [tex]V_{Wx} = 111 \cdot \cos(66^{\circ}) \approx 45.3 \text{ km/h}[/tex]
    • [tex]V_{Wy} = 111 \cdot \sin(66^{\circ}) \approx 103.0 \text{ km/h}[/tex]

Step 4: Adding the Components:

To find the resultant velocity components:

  • X components:
    [tex]V_x = V_{Ax} + V_{Wx} = -339.6 + 45.3 = -294.3 \text{ km/h}[/tex]

  • Y components:
    [tex]V_y = V_{Ay} + V_{Wy} = 616.1 + 103.0 = 719.1 \text{ km/h}[/tex]

Step 5: Resultant Velocity Magnitude:

Now we compute the magnitude of the resultant velocity vector:
[tex]V = \sqrt{V_x^2 + V_y^2}[/tex]
[tex]V = \sqrt{(-294.3)^2 + (719.1)^2} \approx \sqrt{86612.88 + 517176.45} \approx \sqrt{603789.33} \approx 777.0 \text{ km/h}[/tex]

Step 6: Direction of the Resultant Velocity:

To find the direction of the resultant vector, we calculate the angle using:
[tex]\theta = \tan^{-1}\left(\frac{V_y}{V_x}\right)[/tex]
[tex]\theta = \tan^{-1}\left(\frac{719.1}{-294.3}\right) \approx -67.1^{\circ}[/tex]


This means the angle is measured clockwise from the negative x-axis (west), giving us a directional angle in relation to east.

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