Question

A jet airliner moving initially at 3.00 3 102 mi/h due east enters a region where the wind is blowing 1.00 3 102 mi/h in a direction 30.0° north of east. (a) Find the components of the velocity of the jet air- liner relative to the air, SvJA. (b) Find the components of the velocity of the air relative to Earth,

Answer 1

The components of the velocity of the jet airliner relative to the air are 3.86 x 10^2 mi/h due east and 5.00 x 10^1 mi/h north. The components of the velocity of the air relative to Earth are 2.13 x 10^2 mi/h due east and -5.00 x 10^1 mi/h south.

Step-by-step explanation:

(a) To find the components of the velocity of the jet airliner relative to the air, we can use vector addition. The velocity of the jet airliner relative to the air is the sum of the velocity of the jet airliner relative to the ground and the velocity of the wind relative to the ground. Since the jet airliner is moving due east and the wind is blowing in a direction 30.0° north of east, we can break down the velocities into their components:

• The velocity of the jet airliner relative to the ground is 3.00 x 10^2 mi/h due east, so its x-component is 3.00 x 10^2 mi/h and its y-component is 0 mi/h.
• The velocity of the wind relative to the ground is 1.00 x 10^2 mi/h in a direction 30.0° north of east, so its x-component is 1.00 x 10^2 mi/h * cos(30.0°) = 8.66 x 10^1 mi/h and its y-component is 1.00 x 10^2 mi/h * sin(30.0°) = 5.00 x 10^1 mi/h.

Adding the x-components and y-components separately, we get the components of the velocity of the jet airliner relative to the air as:

• x-component: 3.00 x 10^2 mi/h + 8.66 x 10^1 mi/h = 3.86 x 10^2 mi/h due east
• y-component: 0 mi/h + 5.00 x 10^1 mi/h = 5.00 x 10^1 mi/h north

Therefore, the components of the velocity of the jet airliner relative to the air are 3.86 x 10^2 mi/h due east and 5.00 x 10^1 mi/h north.

(b) To find the components of the velocity of the air relative to Earth, we can use vector subtraction. The velocity of the air relative to Earth is the difference between the velocity of the jet airliner relative to the ground and the velocity of the wind relative to the ground. Since the velocity of the jet airliner relative to the ground is already known, we can simply subtract the components of the velocity of the wind relative to the ground:

• The x-component of the velocity of the air relative to Earth is the x-component of the velocity of the jet airliner relative to the ground minus the x-component of the velocity of the wind relative to the ground: 3.00 x 10^2 mi/h - 8.66 x 10^1 mi/h = 2.13 x 10^2 mi/h due east.
• The y-component of the velocity of the air relative to Earth is the y-component of the velocity of the jet airliner relative to the ground minus the y-component of the velocity of the wind relative to the ground: 0 mi/h - 5.00 x 10^1 mi/h = -5.00 x 10^1 mi/h south.

Therefore, the components of the velocity of the air relative to Earth are 2.13 x 10^2 mi/h due east and -5.00 x 10^1 mi/h south.

Answer 2

a) Vaw = 2.19 mi / h θ = 346.8º b) Vmgx = 0.866 mi / h Vawy = -0.5 mi / h

Step-by-step explanation:

We using the sum of vectors, we use the indexes ‘a’ for the plane, ‘g’ for the land ‘w’ for the wind,

`v_(ag)` =
`v_(aw)` +
`v_(wg)`

`v_(aw)` =
`v_(ag)` +
`v_(wg)`

To facilitate the calculation we decompose with respect to xy coordinate system

`v_(wgx)` =
`v_(wg)` cos 30

`v_(wgy)`=
`v_(wg)` sin30

Vwgx = Vwd cos 30

Vwgx = Vwd sin30

Vmgx = 1.00 cos 30

Vmgx = 0.866 mi / h

Vmgy = 1.00 sin30

Vwgy = 0.5 mi / h

Let's find the resulting components

Vawx = 3.00 -0.866

Vawx = 2,134 mi / h

Vawy = 0 - 0.5

Vawy = -0.5 mi / h

Let's use Pythagoras' theorem

Vaw2 = Vawx2 + Vawy2

Vaw = Ra Vawx2 + Vawy2

Vaw = ra 2,134 2 + 0.52

Vaw = 2.19 mi / h