a plane is flying due north at 155 mi/h and encounters a wind from the west at 45m/h. what is the plane's new velocity with respect to the ground in standard position.

Question

a plane is flying due north at 155 mi/h and encounters a wind from the west at 45m/h. what is the plane's new velocity with respect to the ground in standard position.

answer is 128 mi/h at 239.4. how do i get to this answer.

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3 answers
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Answer 1

Lisa answered
14 years ago

a plane is flying southwest at 155 mi/h and encounters a wind from the west at 45m/h. what is the plane's new velocity with respect to the ground in standard position.

answer is 128 mi/h at 239.4. how do i get to this answer.

Answer 2

drwls answered
14 years ago

I do not agree with that answer. You should add 155 N and 45 W as vectors.

I get a ground speed of 161.4 mph, with a direction 16.2 degrees West of North.

Answer 3

Explain Bot answered
11 months ago

To find the plane's new velocity with respect to the ground, we can use vector addition. The velocity of the plane will have two components: the velocity due to its own motion (due north at 155 mi/h) and the velocity due to the wind (from the west at 45 mi/h).

1. Convert the velocity components into vector form:
- The velocity due to the plane's own motion can be represented as (0, 155) mi/h.
- The velocity due to the wind can be represented as (-45, 0) mi/h since it is from the west and there is no component in the north-south direction.

2. Add the two vectors together to find the resultant velocity:
- To add the vectors, simply add their corresponding components. In this case, add the x-components and the y-components separately.
- (0 + (-45), 155 + 0) = (-45, 155) mi/h

3. Find the magnitude and direction of the resultant velocity:
- The magnitude of the resultant velocity can be calculated using the Pythagorean theorem: √((-45)^2 + 155^2) = √(2025 + 24025) = √26050 ≈ 161.35 mi/h.
- The direction can be found using trigonometry. Since the resultant velocity has a negative x-component and a positive y-component, the angle will be in the second quadrant. The inverse tangent of the y-component divided by the x-component gives the angle: arctan(155/-45) ≈ -74.22 degrees.
- Convert the angle to the standard position by adding 180 degrees to get 180 - 74.22 ≈ 105.78 degrees.

Therefore, the plane's new velocity with respect to the ground is approximately 161.35 mi/h at 105.78 degrees (measured counterclockwise from the positive x-axis).