Question

NEED HELP ASAP PLEASE!!!
An airplane is headed due south with an airspeed of 300 kph. A wind is blowing from the west at 120 kph. What is the ground speed and direction of the plane?

323 kph with a directional bearing of S 22° E
328 kph with a directional bearing of S 68° E
420 kph with a directional bearing of S 22° E
420 kph with a directional bearing of S 68° E

Answer 1

Answer:

a) 323 kph with a directional bearing of S 22° E

Step-by-step explanation:

to find the ground speed:

just do the pythagorean theorem.

c²=300²+120²

c²=104400

c=323

to find the directional bearing:

use the law of sines!

sin(90)/323 = sin(θ)/120

(*explanation: since the plane is heading due south and the wind is from due west, it makes a 90 degree angle; and the side opposite of the 90 degree angle is 323. set it equal to sin theta over 120; theta is the angle u r trying to find, and the side opposite of it is 120)

cross multiply:

120sin90=323sin(θ)

divide 323 on both sides:

sin(θ)=0.3715

use inverse sin to find out degree measure:

sin⁻¹(0.3715)=21.8

rounds up to 22.

Answer 2

The ground speed and direction of the plane is 323 kph with a directional bearing of S 22° E. Therefore, option A is the correct answer.

To find the ground speed of the plane, we need to know what a vector is

What is a vector?

A vector is a variable that has both magnitude and direction.

Since the airplane is headed due south with an airspeed of 300 kph, its vector is V = (-300 kph)j.

Also, the wind is blowing from the west at 120 kph. Its vector is v = (-120 kph)i

Now, the ground speed of the plane V' is the resultant vector of the airspeed and wind speed.

What is a resultant vector?

A resultant vector is the sum of two or more vectors.

So, V' = v + V

V' = (-120 kph)i + (-300 kph)j

So, its magnitude V' = √(x² + y²) where

x = -120 kph and

y = -300 kph.

So, V' = √[(-120 kph)² + (-300 kph)²]

= √[14400 kph² + 90000 kph²]

= √(104400 kph²)

= 323.11 kph

≅ 323 kph

The direction of the plane

Its direction, Ф = tan⁻¹(y/x)

Ф = tan⁻¹(-300 kph/-120 kph)

Ф = tan⁻¹(2.5)

Ф = 68.2°

From the South-line we have α = 90° - Ф = 90° - 68.2° = 21.8° ≅ 22°

So, the directional bearing of the plane is S22°E.

So, the ground speed and direction of the plane is 323 kph with a directional bearing of S 22° E.

Learn more about the ground speed of a plane and direction here:

brainly.com/question/11861018

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