Solving Vectors Word Problem: Ground Velocity of Airplane

[physics4ever25]

Excellent. Now, can you use trigonometry (with the angles 50 degrees and 30 degrees for the two vectors) to get the x and y components of each of the two vectors?

For the wind vector, the x component is 129.90 km/h and the y component is 75 km/h. For the plane vector, the x component is 321.39 km/h and the y component is 383.02 km/h.

 

[Chestermiller]

For the wind vector, the x component is 129.90 km/h and the y component is 75 km/h. For the plane vector, the x component is 321.39 km/h and the y component is 383.02 km/h.

The magnitudes of these components are correct, but you already said that the sign of the x component of the wind is negative. So, what is the x component of the wind, with the correct sign?

 

[physics4ever25]

The magnitudes of these components are correct, but you already said that the sign of the x component of the wind is negative. So, what is the x component of the wind, with the correct sign?

It would be -129.90 km/h. At this point I would total the x components and total the y components to get a total of 191.49 km/h for the x components and a total of 458.02 km/h for the y components. This will form a triangle (for the totals) that will look as follows:
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The remaining side will be the resultant so I will use Pythagorean theorem to find it, so-
r=square root (458.02^2+191.49^2)
r=496.44

Therefore, the magnitude of the resultant will be 496.44 km/h. Now, in terms of the angle, I don't know which of the two angles will be the one.

 

[Chestermiller]

It would be -129.90 km/h. At this point I would total the x components and total the y components to get a total of 191.49 km/h for the x components and a total of 458.02 km/h for the y components. This will form a triangle (for the totals) that will look as follows:
____________________________________________________________________________________________________________________________________

View attachment 96259

____________________________________________________________________________________________________________________________________

The remaining side will be the resultant so I will use Pythagorean theorem to find it, so-
r=square root (458.02^2+191.49^2)
r=496.44

Therefore, the magnitude of the resultant will be 496.44 km/h. Now, in terms of the angle, I don't know which of the two angles will be the one.

What is the angle that the resultant makes with the negative x axis?

 

[physics4ever25]

Would it be the one on the lower right?

 

[Chestermiller]

Would it be the one on the lower right?

I don't know what this means. I get an angle of about 67.4 degrees between the resultant and the negative x axis.

 

[physics4ever25]

I don't know what this means. I get an angle of about 67.4 degrees between the resultant and the negative x axis.

Yes, I got 67.3 degrees.

 

[Chestermiller]

Yes, I got 67.3 degrees.

OK. Now going around clockwise from North, the negative x-axis is at a bearing of 270 degrees (due West). That would be at 9 o'clock. But the resultant vector is even further clockwise than this, more towards 11 o'clock. So what is the bearing of the resultant vector?

Chet

 

[physics4ever25]

Would it be 337.3 degrees?

 

[Chestermiller]

Would it be 337.3 degrees?

Sure. Now we're done. Pretty easy, huh? Want to try another case?

 

[physics4ever25]

Sure. Now we're done. Pretty easy, huh? Want to try another case?

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):

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While on a search and rescue mission for a boat lost at sea, a helicopter leaves its pad and travels 75 km at N20E, turns and travels 43 km at S70E, turns again and travels 50 km at S24W, and makes a final turn and travels 18 km at N18W, where the boat is found. What is the displacement of the boat from the helicopter pad?

_______________________________________________________________________________________________________________________________________

 

[Chestermiller]

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):

_______________________________________________________________________________________________________________________________________

While on a search and rescue mission for a boat lost at sea, a helicopter leaves its pad and travels 75 km at N20E, turns and travels 43 km at S70E, turns again and travels 50 km at S24W, and makes a final turn and travels 18 km at N18W, where the boat is found. What is the displacement of the boat from the helicopter pad?

_______________________________________________________________________________________________________________________________________

OK. So now you have 4 vectors. First draw your diagram with the tail of each vector at the origin. Then, what is the acute included angle that each vector makes with the x axis?

 

[physics4ever25]

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 x-axis; vector 3 makes an angle of 66 degrees with the x-axis; and vector 4 makes an angle of 72 degrees with the x-axis.

 

[Chestermiller]

View attachment 96265

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 x-axis; vector 3 makes an angle of 66 degrees with the x-axis; and vector 4 makes an angle of 72 degrees with the x-axis.

The angle for vector 2 is incorrect. Also, I recommend plotting each of the vectors with its tail at the origin. This will make it easier to figure out the components.

 

[physics4ever25]

The angle for vector 2 is incorrect. Also, I recommend plotting each of the vectors with its tail at the origin. This will make it easier to figure out the components.

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)?

 

[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