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A plane traveling at 300 mph is flying with a bearing of 30°. There is a wind of 50 mph from the south. If no correction is made for the wind, what are the final bearing and ground speed of the plane?

Respuesta :

oladayoademosu

The final bearing and ground speed of the plane are 21.05° and 278.39 mph respectively.

Since the plane is traveling at 300 mph flying with a bearing of 30°, it will have both horizontal and vertical components of velocity.

The vertical component of velocity v = 300sin30° = 300 × 0.5 = (150 mph)j and the horizontal component v' = 300cos30° = 300 × 0.8660 = (259.81 mph)i.

Now, since the wind blows 50 mph south, its direction is in the negative y direction. So, v" = -(50 mph)j in the y direction

So, the plane's ground velocity = velocity of plane + velocity of wind

Now, the vertical component of the plane's ground speed V = vertical component of velocity of plane + velocity of wind

= (150 mph)j + (- 50 mph)j

= (150 mph)j - (50 mph)j

= (100 mph)j

The horizontal component of the velocity of the plane's ground speed V' equals the horizontal component of the velocity of the plane.

V' = (259.81 mph)i

So, the plane's ground speed V" = resultant of V and V'

= √(V² + V'²)

= √(100² + 259.81²)

= √(10000 + 67501.2361)

= √77501.2361

= 278.39 mph

The plane's bearing relative to the ground is Ф = tan⁻¹(V/V')

= tan⁻¹(100/259.81)

= tan⁻¹(0.3849)

= 21.05°

So, the final bearing and ground speed of the plane are 21.05° and 278.39 mph respectively.

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adioabiola

We are required to find the final bearing and ground speed of the plane

The final bearing of the plane is : π = 37.59°

And, the final ground speed, Vfinal = 327.87mph.



The plane travelling at 300mph with a bearing of 30° (with the horizontal).....

Therefore, the horizontal component of the plane's speed = Vx = VCos30°

And, the vertical component of the plane's speed = Vy = VSin30°

Therefore, Vx = 300× 0.866 = 259.8mph

And, Vy = 300× 0.5 = 150mph

If there's a wind Vw = 50mph from the south, this means the wind is going In the positive y direction.

Therefore, the total horizontal component of the speed, Vyy is the algebraic sum of Vy and Vw.

Vyy = (150 + 50) mph.

Therefore, the final bearing of the plane is π = (tan⁻¹ Vyy/Vx)

π = (tan⁻¹ 200/259.8)

π = 37.59°

And the final ground speed, Vfinal = √(200² + 259.8²)

Therefore, the final ground speed, Vfinal = 327.87mph.

Read more:

https://brainly.com/question/18228407

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