Recent content by physics4ever25

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.

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

Ah, that makes more sense, but just confused about something- wouldn't both the previous question and this question fall in quadrant 1 then? I mean, the total x and total y components for both the previous question and this question were positive.

I know that but I don't know how to figure out the resultant angle. I mean, in order to find the resultant angle you will need to first determine what quadrant the resultant vector is located in. Once you have determined the quadrant it is in, then you can go about figuring out the resultant...

I mean in general I don't know how to find the direction of the resultant. I know how to find the missing angle in the triangle (using tan theta) but I don't know how to find the actual resultant angle (bearing).

That's what I'm confused about. How do you determine that the resultant vector was pointing at 11 O clock for the previous question and at 2 O clock for this question.

I'm a bit confused by how you found the hand on the clock. For the previous question you determined that it was around 11 and for this one you determined that it's around 2. However, I don't know how to determine this.

The acute angle would be 34.12 degrees. I don't know how to find the true bearing or the direction in terms of quadrant bearing.

Total x's would be 40.16 and the total y's would be 27.21. Thus, the magnitude of the the resultant would be 48.51 (got this using Pythagorean theorem).

For vector 1, the x is 25.65 and y is 70.48; for vector 2, the x is 40.41 and y is -14.71; for vector 3, the x is -20.34 and y is -45.68; and for vector 4, the x is -5.56 and y is 17.12.

For vector 1, x and y both are positive; for vector 2, x is positive and y is negative; for vector 3, x and y both are negative; for vector 3, x is negative and y is positive.

My bad, the angle for vector 2 (in relation to the x-axis) is actually 20 degrees. Also, just curious, why are we shifting the vectors such that the tail is at the origin (for this question and for the previous one)?

This is very messy, so it may be hard to see what's what. ____________________________________________________________________________________________________________________________________ Vector 1 makes an angle of 70 degrees with the x-axis; vector 2 makes an angle of 47 degrees with the...

Yeah, that wasn't as bad as I thought it was. Yes, sure, let's try another case. Here's another question (not plane/wind but it involves the same concept of vectors)...

Would it be 337.3 degrees?