Vectors - Groundspeed question

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A plane is flying N45°E at an airspeed of 525 km/h, with wind coming from N60°W at 98 km/h. The initial calculations using the cosine law yielded a ground speed of 483.5 km/h and an angle of 10.1°. After adjusting for the correct angle due to wind direction, the resultant ground speed was recalculated to 558.4 km/h, with a new angle of approximately 9.8°. The final confirmations indicate that the calculations align closely, affirming the method used.
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Homework Statement


A plane is steering N45°E with an airspeed of 525 km/h. The wind is from N60°W at 98 km/h. Find the ground speed and the direction of the plane.


Homework Equations


Cosine law, sine law, vector addition/subtraction.


The Attempt at a Solution


I have attached an image of my diagram (sorry for the messiness!)

I Solved for R using the cosine law:

R^2 = 525^2 + 98^2 - 2(525)(98)cos60°
R^2 = 285229 - 102900cos60
R^2 = 233779
R = 483.5 km/h

As for the angle, theta, I also used the cosine law:

98^2 = 525^2 + 483.5^2 = 2(525)(483.5)cosθ
-499793.25 = -507675cosθ
cosθ = 0.9844
θ = 10.1°

Is my method correct?

Thank you in advance!
 

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kariibex said:

Homework Statement


A plane is steering N45°E with an airspeed of 525 km/h. The wind is from N60°W at 98 km/h. Find the ground speed and the direction of the plane.

Homework Equations


Cosine law, sine law, vector addition/subtraction.

The Attempt at a Solution


I have attached an image of my diagram (sorry for the messiness!)

I Solved for R using the cosine law:

R^2 = 525^2 + 98^2 - 2(525)(98)cos60°
R^2 = 285229 - 102900cos60
R^2 = 233779
R = 483.5 km/h

As for the angle, theta, I also used the cosine law:

98^2 = 525^2 + 483.5^2 = 2(525)(483.5)cosθ
-499793.25 = -507675cosθ
cosθ = 0.9844
θ = 10.1°

Is my method correct?

Thank you in advance!
attachment.php?attachmentid=58446&d=1367618081.png


The wind is from N60°W ...

That means the angle you label as 60° should be 60° + 45°.
 
SammyS said:
attachment.php?attachmentid=58446&d=1367618081.png


The wind is from N60°W ...

That means the angle you label as 60° should be 60° + 45°.

Thank you for the reply.

Okay, so using 60° + 45°, I calculated my new resultant vector as 558.4 km/hr.

The only issue is when I solve for theta, cosθ = 1.138, which cannot be solved for.

Is there something I'm missing?
 
kariibex said:
Thank you for the reply.

Okay, so using 60° + 45°, I calculated my new resultant vector as 558.4 km/hr.

The only issue is when I solve for theta, cosθ = 1.138, which cannot be solved for.

Is there something I'm missing?
What is the equation you have when solving for cos(θ)?
 
SammyS said:
What is the equation you have when solving for cos(θ)?

Oh! I'm sorry, I just realized my mistake (substituted the wrong value) :)

Anyways, my new θ turns out to be:

98^2 = 525^2 + 558.4^2 - 2(525)(558.4)cosθ
-577831.5 = -586320cosθ
cosθ = 0.9855
θ = 9.8°

Hopefully my method is correct?

Thank you for all your help, I really appreciate it :)
 
kariibex said:
Oh! I'm sorry, I just realized my mistake (substituted the wrong value) :)

Anyways, my new θ turns out to be:

98^2 = 525^2 + 558.4^2 - 2(525)(558.4)cosθ
-577831.5 = -586320cosθ
cosθ = 0.9855
θ = 9.8°

Hopefully my method is correct?

Thank you for all your help, I really appreciate it :)
Yup. I got 9.759° .
 
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