Question

You are on an airplane traveling 30° south of due west at 140 m/s with respect to the air. The air is moving with a speed 35 m/s with respect to the ground due north.

1) What is the speed of the plane with respect to the ground?
2) What is the heading of the plane with respect to the ground? (Let 0° represent due north, 90° represents due easth).
3) How far west will the plane travel in 1 hour?

Answer 1

1) The speed of the plane with respect to the ground is 143.6 m/s. 2) The heading of the plane with respect to the ground is 157.2°. 3) The plane will travel 140.6 km west in 1 hour.

Step-by-step explanation:

1) To find the speed of the plane with respect to the ground, we need to calculate the resultant velocity. The velocity of the plane with respect to the ground is the vector sum of the velocity of the plane with respect to the air and the velocity of the air with respect to the ground. We can use the concept of vector addition to calculate this.

Let's break down the given information:

Velocity of the plane with respect to the air: 140 m/s at an angle 30° south of due west

Velocity of the air with respect to the ground: 35 m/s due north

To find the resultant velocity, we can use the law of cosines:

Resultant velocity = sqrt((v_plane_air)^2 + (v_air_ground)^2 - 2(v_plane_air)(v_air_ground)cos(angle_between_vectors))

Plugging in the values, we get:

Resultant velocity = sqrt((140)^2 + (35)^2 - 2(140)(35)cos(180° - 30°))

Resultant velocity = 143.6 m/s

Therefore, the speed of the plane with respect to the ground is 143.6 m/s.

2) To find the heading of the plane with respect to the ground, we need to calculate the angle between the direction of the resultant velocity and the due north direction. We can use the inverse tangent function to calculate this:

Heading = arctan((v_plane_air sin(angle_between_vectors))/(v_plane_air cos(angle_between_vectors) + v_air_ground)

Plugging in the values, we get:

Heading = arctan((140 sin(180° - 30°))/(140 cos(180° - 30°) + 35))

Heading = 157.2°

Therefore, the heading of the plane with respect to the ground is 157.2°.

3) To calculate how far west the plane will travel in 1 hour, we need to multiply the westward component of the velocity of the plane with respect to the ground by the time. The westward component is given by:

Westward component = v_plane_ground sin(heading)

Plugging in the values, we get:

Westward component = 143.6 sin(157.2°)

Westward component = -140.6 m/s

The negative sign indicates the direction west. Since the plane is traveling west, we can ignore the negative sign.

Therefore, the plane will travel 140.6 km west in 1 hour.

Answer 2

1 ) By the Cosine Law:
v² = 140² + 35² - 2 · 140 · 35 · cos 60°
v² = 19,600 + 1,225 - 2 · 4,900 · 1/2 = 15,925
v = √15,925
v = 126.2 m/s
2 ) By the Sine Law:
126.2 / sin 60° = 35 / sin x
126.2 / 0.866 = 35/sin x
sin x = 0.24
x = sin^(-1) 0.24 = 13.9°
α = 30° - 13.9° = 16.1°
The heading of the plane ( 270° represents due West ):
270° - 16.1° = 253.9°
3 ) In 1 hour:
d = 126.2 m/s · 3600 s = 454,320 m
454,320 m · cos 16.1° = 454,320 m · 0.96 = 436,147.2 m