A jet airliner moving initially at 915 mph (with respect to the ground) to the east moves into a region where the wind is blowing at 969 mph in a direction 34 degrees north of east. What is the new speed of the aircraft with respect to the ground? Answer in units of mph.

1 answer

To find the new speed of the aircraft with respect to the ground, we need to perform vector addition of the aircraft's velocity and the wind's velocity.

  1. Set up the vectors:

    • The aircraft's velocity vector (\( \vec{V}{\text{plane}} \)) can be represented as: \[ \vec{V}{\text{plane}} = (915, 0) \] (since it's moving east, there's no northward component).

    • The wind's velocity vector (\( \vec{V}{\text{wind}} \)) at a speed of 969 mph, 34 degrees north of east can be broken down into its components: \[ \vec{V}{\text{wind}} = (969 \cos(34^\circ), 969 \sin(34^\circ)) \]

  2. Calculate the components of the wind’s velocity:

    • We first calculate \( \cos(34^\circ) \) and \( \sin(34^\circ) \):

      • Using a calculator, we find: \[ \cos(34^\circ) \approx 0.8290 \] \[ \sin(34^\circ) \approx 0.5592 \]
    • Now, calculate the components of the wind: \[ V_{\text{wind}, x} = 969 \cdot \cos(34^\circ) \approx 969 \cdot 0.8290 \approx 805.661 \] \[ V_{\text{wind}, y} = 969 \cdot \sin(34^\circ) \approx 969 \cdot 0.5592 \approx 542.229 \]

    • Therefore, the wind velocity vector is approximately: \[ \vec{V}_{\text{wind}} \approx (805.661, 542.229) \]

  3. Add the velocity vectors:

    • Now we add the two velocity vectors: \[ \vec{V}{\text{total}} = \vec{V}{\text{plane}} + \vec{V}_{\text{wind}} = (915, 0) + (805.661, 542.229) = (1720.661, 542.229) \]
  4. Calculate the magnitude of the resultant velocity:

    • The magnitude of the resultant vector is given by: \[ V_{\text{total}} = \sqrt{(1720.661)^2 + (542.229)^2} \]

    • Calculating each component: \[ (1720.661)^2 \approx 2951586.578 \] \[ (542.229)^2 \approx 294805.087 \]

    • Now sum these values: \[ V_{\text{total}} = \sqrt{2951586.578 + 294805.087} \approx \sqrt{3246391.665} \approx 1803.55 \]

  5. Final answer:

    • Therefore, the new speed of the aircraft with respect to the ground is approximately: \[ \boxed{1803.55} \text{ mph} \]
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