Solving Vectors Word Problem: Ground Velocity of Airplane

[Chestermiller]

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)?

We shift the vectors because it makes it easier to visually see the horizontal and vertical components, and the angles.

20 degrees is correct.

Now, for each of the 4 vectors, from what you see in the diagram, what is the sign of the x component and the y component.

 

[physics4ever25]

We shift the vectors because it makes it easier to visually see the horizontal and vertical components, and the angles.

20 degrees is correct.

Now, for each of the 4 vectors, from what you see in the diagram, what is the sign of the x component and the y component.

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.

 

[Chestermiller]

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.

Excellent. Now get the x and y components of each of the 4 vectors, including the correct signs.

 

[physics4ever25]

Excellent. Now get the x and y components of each of the 4 vectors, including the correct signs.

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.

 

[Chestermiller]

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.

These look right. Now, what are the components of the resultant? What is the magnitude of the resultant?

 

[physics4ever25]

These look right. Now, what are the components of the resultant? What is the magnitude of the resultant?

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).

 

[Chestermiller]

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).

OK. What is the acute included angle between this and the x axis? What is the direction in terms of N??E

 

[physics4ever25]

OK. What is the acute included angle between this and the x axis? What is the direction in terms of N??E

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.

 

[Chestermiller]

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.

The angle is correct. The bearing is determined by the hand on the clock again. It's at an angle of 56 degrees clockwise from North (about 2 o'clock). The quadrant bearing, according to the algorithm implied by the input data is N56E.

You did very well. I don't think you'll have any problems with the component method any more.

Chet

 

[physics4ever25]

The angle is correct. The bearing is determined by the hand on the clock again. It's at an angle of 56 degrees clockwise from North (about 2 o'clock). The quadrant bearing, according to the algorithm implied by the input data is N56E.

You did very well. I don't think you'll have any problems with the component method any more.

Chet

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.

 

[Chestermiller]

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.

If the resultant vector were one of the hands on a clock, with 12 at North and 6 at South, 3 at East, and 9 at West, at what time would the resultant vector be pointing?

 

[physics4ever25]

If the resultant vector were one of the hands on a clock, with 12 at North and 6 at South, 3 at East, and 9 at West, at what time would the resultant vector be pointing?

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.

 

[Chestermiller]

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.

If the clock analogy doesn't work for you, forget about it. It was just another way that I thought of for looking at the direction of the resultant. It's not worth spending your valuable time trying to figure it out.

 

[physics4ever25]

If the clock analogy doesn't work for you, forget about it. It was just another way that I thought of for looking at the direction of the resultant. It's not worth spending your valuable time trying to figure it out.

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).

 

[Chestermiller]

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).

The bearing is the angle between the resultant vector and a vector pointing due north, with the resultant vector angle measured clockwise from due north. Due north is 0 degrees bearing. Due east is 90 degrees bearing. Due south is 180 degrees bearing. Due west is 270 degrees bearing.

 

[physics4ever25]

The bearing is the angle between the resultant vector and a vector pointing due north, with the resultant vector angle measured clockwise from due north.

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 angle. I don't know how to determine which quadrant it is located in. Like, for the previous question the resultant vector was located in quadrant 1 and for this question the resultant vector is located in quadrant 2. How do I determine this?

 

[Chestermiller]

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 angle. I don't know how to determine which quadrant it is located in.

In quadrant 1, the x and y components are both positive. The bearing is 0 to 90 degrees.
In quadrant 2, the x component is negative and the y component is positive. The bearing is 270 to 360 degrees.
In quadrant 3, the x component is negative and the y component is negative. The bearing is 180 to 270 degrees.
In quadrant 4, the x component is positive and the y component is negative. The bearing is 90 to 180 degrees.

The quadrants are numbered in counter-clockwise order. The bearings are numbered in clockwise order.

 

[physics4ever25]

In quadrant 1, the x and y components are both positive. The bearing is 0 to 90 degrees.
In quadrant 2, the x component is negative and the y component is positive. The bearing is 270 to 360 degrees.
In quadrant 3, the x component is negative and the y component is negative. The bearing is 180 to 270 degrees.
In quadrant 4, the x component is positive and the y component is negative. The bearing is 90 to 180 degrees.

The quadrants are numbered in counter-clockwise order. The bearings are numbered in clockwise order.

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.

 

[Chestermiller]

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.

In the previous question, the x component was negative and the y component was positive. So that was 2nd quadrant.

 

[physics4ever25]

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

 

[Chestermiller]

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).

 

[physics4ever25]

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.

 

[aardwolf.sg]

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.