Vector Addition and Bearing Calculation in 2D Motion

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SUMMARY

The forum discussion centers on a Mechanics vector problem involving a particle with two velocities: 4 m/s east and 5 m/s at 30 degrees to the east. The user calculated the resultant vector's magnitude as 8.7 m/s and derived a bearing of 73.3 degrees. However, the textbook states the bearing as 20.5 degrees, leading to confusion. Other participants confirmed the user's calculations, indicating the textbook's answer is incorrect.

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  • Understanding of vector addition in 2D motion
  • Familiarity with trigonometric functions, particularly sine and cosine
  • Knowledge of bearing calculations in navigation
  • Proficiency with scientific calculators for trigonometric computations
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Students studying Mechanics, physics enthusiasts, and educators looking to clarify vector addition and bearing calculations in 2D motion.

essaichay
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Hi,

I have a Mechanics vector problem here split in two parts. First to solve the magnitude of the resultant vector and then to give the bearing at which it's acting at. Worked out most of it, but the last answer to the question isn't the same as mine. Kind of difficult to explain the question since I would need a diagram but hey, here it is:

Question:

A particle has velocity 4ms^-1 acting on it, in the east direction followed by a velocity of 5ms^1 acting at 30 degrees to the east direction.

I found the resultant vector to be of magnitude 8.7 ms^-1, But a bearing of 73.3 degrees?

The magnitude part is right however, they give a bearing of 20.5 degrees rather than 73.3.

For my bearing I done: 90 - tan^-1(5sin30/(4+5cos30)).

Any help? Thanks.
 
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essaichay said:
Hi,

I have a Mechanics vector problem here split in two parts. First to solve the magnitude of the resultant vector and then to give the bearing at which it's acting at. Worked out most of it, but the last answer to the question isn't the same as mine. Kind of difficult to explain the question since I would need a diagram but hey, here it is:

Question:

A particle has velocity 4ms^-1 acting on it, in the east direction followed by a velocity of 5ms^1 acting at 30 degrees to the east direction.

I found the resultant vector to be of magnitude 8.7 ms^-1, But a bearing of 73.3 degrees?

The magnitude part is right however, they give a bearing of 20.5 degrees rather than 73.3.

For my bearing I done: 90 - tan^-1(5sin30/(4+5cos30)).

Any help? Thanks.

"For my bearing I done" is atrocious grammar. For your bearing I did exactly the same thing you did. And I got 73.29468619399003 degrees. Check your calculator skills. The basic math seems fine.
 
Dick said:
"For my bearing I done" is atrocious grammar.

Atrocious grammar? How should it be corrected?

Also, I don't understand how the textbook can give a wrong value of bearing 20.5 degrees when you've ended up with an identical answer to me of 73.3 degrees.

I've checked my calculator skills, they seem pretty fine.
 
Last edited:
It's simple subject-verb agreement. The error itself is inexcusable, but you not seeing it after it was pointed out to you is even worse.
 
essaichay said:
Atrocious grammar? How should it be corrected?

Also, I don't understand how the textbook can give a wrong value of bearing 20.5 degrees when you've ended up with an identical answer to me of 73.3 degrees.

I've checked my calculator skills, they seem pretty fine.

wotcher essaichay! :smile:

It's "For my bearing I done, guv!"

And I make it 73.3 also … looks like the book answer is wrong. :frown:
 
Vid said:
It's simple subject-verb agreement. The error itself is inexcusable, but you not seeing it after it was pointed out to you is even worse.

You know, you could've just replied with a simple correction to the question I asked on how to correct it. "I'm sorry" my grammar isn't the best, but at least you understand (I hope) what I'm saying, and right now, I'm completely confused.

Another option was to actually answer the real query I had about the textbook giving a bearing that was wrong to my solution. And that, I would've been grateful for.
 
tiny-tim said:
wotcher essaichay! :smile:

It's "For my bearing I done, guv!"

And I make it 73.3 also … looks like the book answer is wrong. :frown:

lol. Thank you.
 
Dick and Vid, the internet is world-wide. It's your good fortune that most of it is in a language that you speak fluently. For a lot of the people who post here, English is not their native tongue. Cut them a little slack. I'm sure they will be glad of your correcting their grammar, but not just saying it is wrong.

Essaichay, "done" is the "past participle" of "do". It is always used with "have" or "had". What you wanted was "For my bearing I did ...". Unfortunately, "I done" is so common among English speakers who are just careless with grammar- and other things- I'm afraid it triggered an automatic reaction in Dick.
(And both Dick and Vid did answer your original question.)
 
HallsofIvy said:
Essaichay, "done" is the "past participle" of "do". It is always used with "have" or "had".

yeh … but no … but … http://en.wikipedia.org/wiki/She_Done_Him_Wrong
She Done Him Wrong was nominated for an Academy Award for Best Picture. At 66 minutes, it is the shortest film ever to be so honored.

:smile: … though, technically, IM'UO, it should be "She Done 'im Wrong"! :smile:
 
Last edited:
  • #10
essaichay said:
Atrocious grammar? How should it be corrected?

Also, I don't understand how the textbook can give a wrong value of bearing 20.5 degrees when you've ended up with an identical answer to me of 73.3 degrees.

I've checked my calculator skills, they seem pretty fine.

The grammar thing was 'just kidding'. I actually carelessly thought that the book said 73.3 and you were getting 20.5. That's why I was suggesting you check how you calculated it. Sorry.
 
  • #11
OK! woah! STOOPPP everyone!

CONCLUSION:

-I should've written: did not done
-The textbook has YET AGAIN given me a wrong answer.

:biggrin: Thanks everyone.
 

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