QuestionApril 23, 2025

An airplane is heading north at an airspeed of 550km/hr, but there is a wind blowing from the southwest at 50km/hr. Round answers to two places. The plane will end up flying square degrees off course The plane's speed relative to the ground will be square km/hr

An airplane is heading north at an airspeed of 550km/hr, but there is a wind blowing from the southwest at 50km/hr. Round answers to two places. The plane will end up flying square degrees off course The plane's speed relative to the ground will be square km/hr
An airplane is heading north at an airspeed of 550km/hr, but there is a wind blowing from the southwest at
50km/hr. Round answers to two places.
The plane will end up flying square  degrees off course
The plane's speed relative to the ground will be square  km/hr

Solution
4.5(302 votes)

Answer

The plane will end up flying 3.46 degrees off course. ### The plane's speed relative to the ground will be 586.42 km/hr. Explanation 1. Determine Wind Components The wind from the southwest has components 50\cos(45^\circ) east and 50\sin(45^\circ) north. Calculate these as 50 \times \frac{\sqrt{2}}{2} = 35.36 km/hr for both east and north. 2. Calculate Resultant Velocity The plane's velocity relative to the ground is the vector sum of its airspeed and wind components. North component: 550 + 35.36 = 585.36 km/hr. East component: 35.36 km/hr. 3. Calculate Speed Relative to Ground Use **Pythagorean theorem**: v = \sqrt{(585.36)^2 + (35.36)^2} to find the resultant speed. 4. Calculate Angle Off Course Use **tangent formula**: \theta = \tan^{-1}\left(\frac{35.36}{585.36}\right) to find the angle off course.

Explanation

1. Determine Wind Components<br /> The wind from the southwest has components $50\cos(45^\circ)$ east and $50\sin(45^\circ)$ north. Calculate these as $50 \times \frac{\sqrt{2}}{2} = 35.36$ km/hr for both east and north.<br /><br />2. Calculate Resultant Velocity<br /> The plane's velocity relative to the ground is the vector sum of its airspeed and wind components. North component: $550 + 35.36 = 585.36$ km/hr. East component: $35.36$ km/hr.<br /><br />3. Calculate Speed Relative to Ground<br /> Use **Pythagorean theorem**: $v = \sqrt{(585.36)^2 + (35.36)^2}$ to find the resultant speed. <br /><br />4. Calculate Angle Off Course<br /> Use **tangent formula**: $\theta = \tan^{-1}\left(\frac{35.36}{585.36}\right)$ to find the angle off course.
Click to rate:

Similar Questions