Question 1171107: A commercial jet travels from Miami to Seattle. The jet's velocity with respect to the air is 585 miles per hour, and its bearing is 332°. The jet encounters a wind with a velocity of 50 miles per hour from the southwest.
1. Write the velocity of the wind as a vector in component form.
2. Write the velocity of the jet relative to the air in component form.
3. What is the speed of the jet with respect to the ground? (Round your answer to one decimal place.) (mph)
4. What is the true direction of the jet? (Round your answer to one decimal place.)
N ____ degrees W
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's break down this problem step by step.
**1. Velocity of the Wind as a Vector**
* The wind is from the southwest, meaning it's blowing towards the northeast. This means the wind is at a 45° angle from the horizontal (east) axis, going towards the first quadrant.
* Velocity of the wind = 50 mph.
The components of the wind velocity vector are:
* x-component: 50 * cos(45°) = 50 * (√2 / 2) ≈ 35.355 mph
* y-component: 50 * sin(45°) = 50 * (√2 / 2) ≈ 35.355 mph
Therefore, the velocity of the wind vector is approximately (35.355, 35.355).
**2. Velocity of the Jet Relative to the Air**
* Velocity of the jet = 585 mph
* Bearing = 332°
To find the components, we need to convert the bearing to a standard angle (counterclockwise from the positive x-axis). Since a bearing of 332° means it's 332° clockwise from north, we can find the standard angle by:
* Standard angle = 360° - 332° + 90° = 118°
The components of the jet's velocity vector are:
* x-component: 585 * cos(118°) ≈ -273.74 mph
* y-component: 585 * sin(118°) ≈ 516.33 mph
Therefore, the velocity of the jet relative to the air vector is approximately (-273.74, 516.33).
**3. Speed of the Jet with Respect to the Ground**
To find the jet's velocity with respect to the ground, we add the velocity of the wind to the velocity of the jet relative to the air:
* Ground velocity vector = (35.355 - 273.74, 35.355 + 516.33) ≈ (-238.385, 551.685)
The speed of the jet with respect to the ground is the magnitude of this vector:
* Speed = √((-238.385)^2 + (551.685)^2) ≈ √(56827.4 + 304356.3) ≈ √361183.7 ≈ 600.985 mph
Rounded to one decimal place, the speed is 601.0 mph.
**4. True Direction of the Jet**
To find the true direction, we need to find the angle of the ground velocity vector.
* θ = arctan(y-component / x-component) = arctan(551.685 / -238.385) ≈ arctan(-2.314) ≈ -66.6°
Since the x-component is negative and the y-component is positive, the angle is in the second quadrant. We need to add 180° to get the correct angle:
* θ = -66.6° + 180° = 113.4°
To convert this to a bearing, we subtract 90° and take the absolute value of the result:
* Bearing from North = 113.4° - 90° = 23.4°
* Since the x component is negative, the direction is North 23.4 degrees West.
**Answers:**
1. (35.355, 35.355)
2. (-273.74, 516.33)
3. 601.0 mph
4. N 23.4 degrees W
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