Solving Vectors Word Problem: Ground Velocity of Airplane

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In summary, the conversation discussed a word problem involving vectors and two different methods to solve it. One method used vector diagrams and cosine and sine laws to find the magnitude and angle of the resultant vector. The other method involved converting the vectors to polar form using the Pythagorean theorem and tan. The confusion arose when both methods gave different answers, but upon checking online, both answers were found to be correct. The conversation also touched on the concept of "true bearing" and whether the vectors were measured clockwise or counterclockwise from North. The conversation ended with a discussion on finding the x and y components of the vectors and the resultant vector.
  • #71
physics4ever25 said:
In the previous question the x component was +191.49 so it was a positive x-value?

EDIT: Also I just realized that the correct answer for the previous question in my book is 22.7 degrees for the bearing. We got 337.3 degrees, but the correct answer was actually 360-337.3=22.7
You're right. Your figure in post #38 through me off. I thought that the dashed line was the resultant. You're absolutely right. The bearing should have been 22.7 degrees (90 - 67.3).
 
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  • #72
Chestermiller said:
You're right. Your figure in post #38 through me off. I thought that the dashed line was the resultant. You're absolutely right. The bearing should have been 22.7 degrees (90 - 67.3).

The dashed line is the resultant vector though, isn't it? I mean, we found the magnitude of that side using the Pythagorean theorem, so that side must be the resultant.
 
  • #73
If it's headwind the answer is 505.51km/h @075.62 deg. If it's tailwind the answer is 597.209km/h @83.70 deg. The question says wind from bearing 120 deg so it's headwind. So answer is the first one.
 

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