Vectors and Components-Part One

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Homework Help Overview

The discussion revolves around a problem involving vectors and components, specifically calculating the northerly and easterly components of an aircraft's velocity given its speed and heading angle.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between heading angles and their corresponding components using trigonometric functions. There is a focus on the conversion of the heading angle to a non-heading angle and the implications of this conversion on the calculations.

Discussion Status

Participants have engaged in clarifying the correct application of trigonometric functions to find the components of velocity. Some have confirmed their understanding of the problem setup, while others have questioned the accuracy of their calculations and the definitions of angles used.

Contextual Notes

There is mention of a diagram that participants have drawn to assist in visualizing the problem, although the original problem statement lacks specificity regarding the angle definitions. The discussion includes a focus on ensuring clarity in the interpretation of the heading angle as it relates to the calculations.

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Homework Statement



An aircraft is traveling at 200 mi/h on a heading of 38 degrees. Find the northerly and easterly components of its velocity.



Homework Equations


Sine=opposite/hypotenuse
Cosine=adjacent/hypotenuse

The Attempt at a Solution



The problem states that the angle given is a heading. In order to use our equations, it is best to convert to our non-heading angle. I don't know what that's called, but it's 52 degrees. So to find the northerly component, I did 200cos(52) and got 123.1 mi/h. To get the easterly component, I did 200sin(52) and got 157.6 mi/h. Are these the right answers so far? Do I need to convert back to a "heading" answer? How would I do that?


Thanks so much for a helpful and descriptive answer! :smile:
 
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Medgirl314 said:

Homework Statement



An aircraft is traveling at 200 mi/h on a heading of 38 degrees. Find the northerly and easterly components of its velocity.



Homework Equations


Sine=opposite/hypotenuse
Cosine=adjacent/hypotenuse

The Attempt at a Solution



The problem states that the angle given is a heading. In order to use our equations, it is best to convert to our non-heading angle. I don't know what that's called, but it's 52 degrees. So to find the northerly component, I did 200cos(52) and got 123.1 mi/h. To get the easterly component, I did 200sin(52) and got 157.6 mi/h. Are these the right answers so far? Do I need to convert back to a "heading" answer? How would I do that?


Thanks so much for a helpful and descriptive answer! :smile:
A heading in aviation terms is measured clockwise frpm magnetic north. If the airplane is heading 38 degrees, it means it is moving in a direction 38 degrees east of north, that is, in a generally northeast direction. The problem should more specifically state or sketch the heading, often denoted for this case as N 38 E. in which case , the northerly component of the velocity is 200 cos 38 or 200 sin 52, whichever triangle you choose(draw a sketch!).
 
Okay, I think I got all that. :-) I forgot to mention that I already drew a diagram. The problem isn't any more specific than I indicated, unfortunately. Are you stating that my answer for the northerly component is correct?

Thank you! :smile:
 
Medgirl314 said:
Okay, I think I got all that. :-) I forgot to mention that I already drew a diagram. The problem isn't any more specific than I indicated, unfortunately. Are you stating that my answer for the northerly component is correct?

Thank you! :smile:
Your answer is incorrect. The plane is heading 038 degrees per its compass reading at a speed of 200 mph. Its northerly component is 200 cos 38, not 200 cos 52. Please look at your sketch again to confirm this result.
 
I think I see what you're saying. I should have made my sketch larger and neater. I got it backwards, right? It would be 200 sin 52 if I were to use the this angle, correct?
 
Medgirl314 said:
I think I see what you're saying. I should have made my sketch larger and neater. I got it backwards, right? It would be 200 sin 52 if I were to use the this angle, correct?
Correct!
 
So the northerly component is 157.6 mi./h. Thank you! Was the easterly component correct?
 
Medgirl314 said:
So the northerly component is 157.6 mi./h. Thank you! Was the easterly component correct?
Why no the easterly component is incorrect...what should it be ?
 
Is is 123.1 mi./h? Just for clarity, I got the right answer for the northerly component, correct?

Thanks!
 
  • #10
Medgirl314 said:
Is is 123.1 mi./h? Just for clarity, I got the right answer for the northerly component, correct?

Thanks!
Yes both values for the northerly and easterly components are now correct. ( You might want to round them off however at least to the nearest whole number with no decimal point. ).
 
  • #11
So just 158 mi/h and 123 mi/h. Thank you!
 
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