Solving Vectors Word Problem: Ground Velocity of Airplane

[physics4ever25]

Hello guys,

I have a vectors word problem and I found 2 different ways to solve the same problem but I'm getting different answers. Apparently, both answers are correct since I've looked for the answer online and I found both answers from different sources, so I'm really confused now.

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Question: An airplane is flying at 550 km/h on a heading of 080 degrees. The wind is blowing at 60 km/h from a bearing of 120 degrees. Find the ground velocity (resultant vector) of the airplane.

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Method 1: First you would draw the vector diagram. You would draw a y-axis and a x-axis. Then from the origin you would draw the plane vector at an angle (from north) of 80 degrees and this vector would be 550 km/h in magnitude. Then from the origin you would draw the wind vector at an angle (from north) of 120 degrees and this vector would be 60 km/h in magnitude. Then you would connect the outside ends (away from the origin) of each vector together. This side would be the resultant. This will form a triangle. Then using cosine and sine laws you can find the magnitude and angle of the resultant vector.

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Method 2: I found this method online and I don't even quite understand it myself but apparently it works. All values used in this method are from the question.

Vr=Vp+Vw
=[550(cos80+i*sin80)] + [60(cos120+i*sin120)]
=[95.50+541.64i] + [-30+51.96i]
=65.50+593.60i

Since this is in a+bi form you will now convert it to polar form. You will do this using the Pythagorean theorem and tan.

r=square root (65.50^2+593.60^2)
=597.2

TanTheta=593.60/65.50
Theta=83.70

Therefore, the magnitude of the resultant would be 597.2 km/h and the angle would be 83.70 degrees (from north).

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Anyways, these are the two methods. If you solve this using method 1 you will get a different answer than method 2. However, both answers are apparently correct since I checked for the answer to this question online and I found both answers. I'm confused now about which answer would be the correct one (using method 1 or method 2).

 

[Chestermiller]

If the bearing is x degrees, does this mean that it is x degrees measured clockwise relative to North?

 

[A.T.]

Apparently, both answers are correct since I've looked for the answer online and I found both answers from different sources

You think something is correct, because you found it online?

 

[physics4ever25]

@Chester, if it's "true bearing" then it's from North. In this case it was "true bearing," so that's why it's from North.

@A.T, I wouldn't think it's correct if I found it online from 1 source. I checked multiple sources and found the same answer, and that's how I concluded it probably is the right answer.

 

[Ray Vickson]

@Chester, if it's "true bearing" then it's from North. In this case it was "true bearing," so that's why it's from North.

@A.T, I wouldn't think it's correct if I found it online from 1 source. I checked multiple sources and found the same answer, and that's how I concluded it probably is the right answer.

You still left one part of Chestermiller's question unanswered: do you mean 120 degrees clockwise or counterclockwise (from North)?

 

[physics4ever25]

Yes, it's clockwise.

 

[phinds]

Please draw a vector diagram to show exactly what you are talking about.

 

[physics4ever25]

 

[Chestermiller]

View attachment 96207

OK. This is the picture I envisioned. Now, do you know how to resolve these vectors into their components in the x and the y directions? If so, what are the x and y components of each of the vectors?

 

[physics4ever25]

OK. This is the picture I envisioned. Now, do you know how to resolve these vectors into their components in the x and the y directions? If so, what are the x and y components of each of the vectors?

Hi, I actually don't know how to do that since I never used that method where I divided the vectors into their x and y components. As you can see in the picture, there's one triangle, so at this point what I would do is simply find the magnitude of the resultant by using the cosine law. That would be as follows-

r^2=60^2+550^2-2(60)(550)*cos40
r^2=306100-66000*cos40
r^2=306100-50558.93
r^2=255541.07
r=505.51

Therefore, the magnitude of the resultant would be 505.51 km/h. This, of course, is according to this method. Using the other method (which I showed in my initial post), the magnitude of the resultant came out to be 597.2 km/h. That's what my confusion was. The two methods give me different answers. When I checked online, some sources showed that the answer was 505.51 km/h (as found by method 1) and some sources showed that the answer was 597.2 km/h (as found by method 2). So, apparently both answers are right.

 

[phinds]

And do you seriously believe both answers CAN be right?

 

[Chestermiller]

Hi, I actually don't know how to do that since I never used that method where I divided the vectors into their x and y components. As you can see in the picture, there's one triangle, so at this point what I would do is simply find the magnitude of the resultant by using the cosine law. That would be as follows-

r^2=60^2+550^2-2(60)(550)*cos40
r^2=306100-66000*cos40
r^2=306100-50558.93
r^2=255541.07
r=505.51

Therefore, the magnitude of the resultant would be 505.51 km/h. This, of course, is according to this method. Using the other method (which I showed in my initial post), the magnitude came out to be 597.2 km/h. That's what my confusion was. The two methods give me different answers.

I get an angle between the two vectors of 140 degrees if I place the vectors tail-to-tail.

If I take the components of the two vectors in the x and y directions, I get:

Plane:
x component = 550 cos 10
y component = 550 sin 10

Wind:
x component = - 60 cos 30
y component = 60 sin 30

What do you get for the resultant x component?
What do you get for the resultant y component?

 

[physics4ever25]

And do you seriously believe both answers CAN be right?

No both can't. That is the confusion. When I was checking for the answer online, I was finding both answers (550.51 km/h and 597.2 km/h), and that really threw me off.

 

[Chestermiller]

The 597.2 gives the magnitude of the difference between the two vectors. This is what I get when I use the 140 degrees, laying the vectors tail to tail. The 505.51 (which is the correct answer) gives the sum of the two vectors. This is what you got when you used the 40 degrees, which is consistent with laying the vectors tail to head to add them. Doing the problem in terms of the components gives the resultant of the two vectors as 505.51 also. So the 597.2 is the incorrect answer.

 

[physics4ever25]

The 597.2 gives the magnitude of the difference between the two vectors. This is what I get when I use the 140 degrees, laying the vectors tail to tail. The 505.51 (which is the correct answer) gives the sum of the two vectors. This is what you got when you used the 40 degrees, which is consistent with laying the vectors tail to head to add them. Doing the problem in terms of the components gives the resultant of the two vectors as 505.51 also. So the 597.2 is the incorrect answer.

I actually just asked someone about this and they said that there was an error made in the following solution:

____________________________________________________________________________________________________________________________________

Vr=Vp+Vw
=[550(cos80+i*sin80)] + [60(cos120+i*sin120)]
=[95.50+541.64i] + [-30+51.96i]
=65.50+593.60i

Since this is in a+bi form you will now convert it to polar form. You will do this using the Pythagorean theorem and tan.

r=square root (65.50^2+593.60^2)
=597.2

TanTheta=593.60/65.50
Theta=83.70
____________________________________________________________________________________________________________________________________

The error was the "120" degrees in the initial statement "=[550(cos80+i*sin80)] + [60(cos120+i*sin120)]." That 120 degrees should have been 300 degrees. The reason being, due to the context of the question, wind is considered a back bearing since it says in the question "the wind is blowing at 60 km/h FROM a bearing of 120 degrees." Due to the term "from" used in the question, that means wind is considered a back bearing. For that reason the back bearing formula would be applied in this case: back bearing=(180 degrees+bearing)=(180+120)=300. Therefore, the 120 in the formula should be replaced with 300. When I do that the answer works out to be 505.50 km/h. The proper solution can be seen below:

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Vr=Vp+Vw
=[550(cos80+i*sin80)] + [60(cos300+i*sin300)]
=[95.50+541.64i] + [30-51.96i]
=125.50+489.68i

Since this is in a+bi form you will now convert it to polar form. You will do this using the Pythagorean theorem and tan.

r=square root (125.50^2+489.68^2)
=505.51

TanTheta=489.68/125.50
Theta=75.63

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[Chestermiller]

I actually just asked someone about this and they said that there was an error made in the following solution:

____________________________________________________________________________________________________________________________________

Vr=Vp+Vw
=[550(cos80+i*sin80)] + [60(cos120+i*sin120)]
=[95.50+541.64i] + [-30+51.96i]
=65.50+593.60i

Since this is in a+bi form you will now convert it to polar form. You will do this using the Pythagorean theorem and tan.

r=square root (65.50^2+593.60^2)
=597.2

TanTheta=593.60/65.50
Theta=83.70
____________________________________________________________________________________________________________________________________

The error was the "120" degrees in the initial statement "=[550(cos80+i*sin80)] + [60(cos120+i*sin120)]." That 120 degrees should have been 300 degrees. The reason being, due to the context of the question, wind is considered a back bearing since it says in the question "the wind is blowing at 60 km/h FROM a bearing of 120 degrees." Due to the term "from" used in the question, that means wind is considered a back bearing. For that reason the back bearing formula would be applied in this case: back bearing=(180 degrees+bearing)=(180+120)=300. Therefore, the 120 in the formula should be replaced with 300. When I do that the answer works out to be 505.50 km/h. The proper solution can be seen below:

____________________________________________________________________________________________________________________________________

Vr=Vp+Vw
=[550(cos80+i*sin80)] + [60(cos300+i*sin300)]
=[95.50+541.64i] + [30-51.96i]
=125.50+489.68i

Since this is in a+bi form you will now convert it to polar form. You will do this using the Pythagorean theorem and tan.

r=square root (125.50^2+489.68^2)
=505.51

TanTheta=489.68/125.50
Theta=75.63

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This is what I get doing the problem in component form.

Chet

 

[physics4ever25]

This is what I get doing the problem in component form.

Chet

Can you explain a bit about the whole "back bearing" concept I mentioned. I'm a bit confused about when to use back bearing. In the case of the plane vector in the question, the forward bearing was used, and in the case of wind vector in the question, the back bearing was used.

 

[Chestermiller]

Can you explain a bit about the whole "back bearing" concept I mentioned. I'm a bit confused about when to use back bearing. In the case of the plane vector in the question, the forward bearing was used, and in the case of wind vector in the question, the back bearing was used.

The key words are "blowing from." That means that the wind is blowing from the 120 degree direction. You have correctly shown the wind direction in your diagram.

 

[physics4ever25]

The key words are "blowing from." That means that the wind is blowing from the 120 degree direction. You have correctly shown the wind direction in your diagram.

So whenever it says "blowing from" I use the back bearing for that formula (method 2), which in this case was (180+120)=300. And, if, for instance it said the wind is "blowing at" a bearing of 120 degrees then I would use the 120 rather than the 300?

 

[Chestermiller]

So whenever it says "blowing from" I use the back bearing for that formula (method 2), which in this case was (180+120)=300. And, if, for instance it said the wind is "blowing at" a bearing of 120 degrees then I would use the 120 rather than the 300?

Yes. But ask yourself this: how important is it to remember this? How many times are you going to run into wind and plane problems in real life?

 

[physics4ever25]

Yes. But ask yourself this: how important is it to remember this? How many times are you going to run into wind and plane problems in real life?

Not many times. It was just for a test I have coming up this Monday. After that we're moving on to the next unit so this will probably show up on the exam and that's about it.

 

[phinds]

So whenever it says "blowing from" I use the back bearing for that formula (method 2), which in this case was (180+120)=300. And, if, for instance it said the wind is "blowing at" a bearing of 120 degrees then I would use the 120 rather than the 300?

That seems to me to be an unnecessarily messy way to look at it. The bearing give you the angle of your vector and the "on a bearing of" or "from a bearing of" tells you which end to put the arrow on.

 

[physics4ever25]

There's another similar question like this one that I got for homework- A cruise ship's captain sets the ship's velocity to be 26 knots at a heading of 080 degrees. The current is flowing toward a bearing of 153 degrees at a speed of 8 knots. What is the ground velocity of the cruise ship?

I tried to solve this problem both ways- using the components method and using method 2, but I got slightly different answers. Using the components way I got the answer to be 29.4 knots (magnitude of resultant) and 95.1 degrees (angle of resultant). Using the other method I got the magnitude to be 29.4 knots (same as the other one) but the angle to be -84.9 degrees (different than the other one). So I got the same magnitude from both methods but different angles.

This is what I did using method 2:

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Vr=Vp+Vw
=[26(cos80+i*sin80)] + [8(cos153+i*sin153)] (here I used 153 because the current has a forward bearing since it says in the question "the current is flowing TOWARD a bearing of 153 degrees; unlike the previous question where it said "from" thus I had to use the back bearing)
=[4.51+25.61i] + [-7.13+3.63i]
=-2.62+29.24i

Convert to polar form

r=square root of [(-2.62)^2+29.24^2]
r=29.4

TanTheta=29.24/(-2.62)
Theta=-84.9

Therefore, magnitude of resultant is 29.4 knots at a bearing (angle) of -84.9 degrees.

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When I used the other method (components) to solve the problem I got the same magnitude (29.4 knots) but a different angle (95.1 degrees instead of -84.9 degrees like above).

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EDIT: Okay, so I just noticed something. If I do the following operation: 180+(-84.9), I get 95.1 degrees as the answer (which is the same angle as the components method). There was actually another question just like this in my homework where I ended up with a negative number for the "a" value in the form "a+bi." I tried the same thing, doing 180+degrees (degrees that I got for the theta value) and I got the correct answer (which I would get using the components method). So, it seems that whenever there is a negative value in for the "a" value in the form "a+bi," after converting to polar form I must add the theta value (degrees) to 180 to get the correct answer for the angle. I'm confused how this works though.

 

[Chestermiller]

I don't have the energy to go through all your math, but I get 29.3 knots and 95.1 degrees. When applying the component method, I always start out by getting the angles with respect to the x axis. In this case, the angles are + 10 degrees and -63 degrees with respect to the positive x axis.

 

[physics4ever25]

I don't have the energy to go through all your math, but I get 29.3 knots and 95.1 degrees. When applying the component method, I always start out by getting the angles with respect to the x axis. In this case, the angles are + 10 degrees and -63 degrees with respect to the positive x axis.

Yeah, I got the same answer for the angle once I added the theta value (I got) to 180. So, when converting from a+bi form to polar form, sometimes the theta value you get is the correct one and other times you'd have to add 180 to the theta value (you get) to get the correct angle. I'm assuming this has something to do with the "signs" of the "a" and "b" values in the form "a+bi," because whenever they're positive I've noticed that the theta value you get is the correct one, and whenever one of them is negative then you'd have to add 180 to the theta value (you get) to get the correct answer.

 

[Chestermiller]

Yeah, I got the same answer for the angle once I added the theta value (I got) to 180. So, when converting from a+bi form to polar form, sometimes the theta value you get is the correct one and other times you'd have to add 180 to the theta value (you get) to get the correct angle. I'm assuming this has something to do with the "signs" of the "a" and "b" values in the form "a+bi," because whenever they're positive I've noticed that the theta value you get is the correct one, and whenever one of them is negative then you'd have to add 180 to the theta value (you get) to get the correct answer.

I don't do this with complex numbers. I just get the components and work with them. It's bulletproof.

 

[physics4ever25]

Yeah, I suppose your method is better overall because it works with any amount of vectors. The method I've been using (method 2) only works with word problems that have 2 vectors. However, when I get into word problems with 3 or 4 or even more vectors then I can't use that method anymore so I'll have to resort to the components method.

It's just that I have difficulty with the components method.

 

[phinds]

I completely agree w/ chestermiller. That second method just seems weird and prone to problems and as he says, the component method is simple and bulletproof (well, like anything it's subject to arithmetic errors, but hey ... )

 

[Chestermiller]

Yeah, I suppose your method is better overall because it works with any amount of vectors. The method I've been using (method 2) only works with word problems that have 2 vectors. However, when I get into word problems with 3 or 4 or even more vectors then I can't use that method anymore so I'll have to resort to the components method.

It's just that I have difficulty with the components method.

Once you draw a diagram showing a given vector (along with its direction with respect to the x axis), it is very easy and intuitive to resolve the vector into its components in the x and y directions. Propose a sample vector, and I'll help you figure out how to resolve it. After you have done it once, you will have no further trouble.

Chet

 

[physics4ever25]

I guess I'll try to learn the components method as well then. Anyways, here's a sample problem (I already have solved it using method 2 and I got the right answer) that I'd like to solve using the components method:

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Question: An airplane is flying at an airspeed of 500 km/h, on a heading of 040 degrees. A 150 km/h wind is blowing from a bearing of 120 degrees. Determine the ground velocity of the airplane and the direction of flight.

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The diagram:

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I forgot to add the resultant but that would be just the missing side to form the triangle.

 

[Chestermiller]

I guess I'll try to learn the components method as well then. Anyways, here's a sample problem (I already have solved it using method 2 and I got the right answer) that I'd like to solve using the components method:

____________________________________________________________________________________________________________________________________

Question: An airplane is flying at an airspeed of 500 km/h, on a heading of 040 degrees. A 150 km/h wind is blowing from a bearing of 120 degrees. Determine the ground velocity of the airplane and the direction of flight.

____________________________________________________________________________________________________________________________________

The diagram: View attachment 96252
_____________________________________________________________________________________________________________________________________

I forgot to add the resultant but that would be just the missing side to form the triangle.

OK. If you put the tail of the blowing vector at the origin, what acute angle does the 500 km/h vector make with the positive x direction, and what acute angle does the 150 km/hr vector make with the negative x axis?

 

[physics4ever25]

OK. If you put the tail of the blowing vector at the origin, what acute angle does the 500 km/h vector make with the positive x direction, and what acute angle does the 150 km/hr vector make with the negative x axis?

If I put the tail of the wind vector at the origin, then it makes an acute angle of 30 degrees with the negative x-axis, and the plane vector will make an acute angle of 50 degrees with the positive x-axis.

 

[Chestermiller]

If I put the tail of the wind vector at the origin, then it makes an acute angle of 30 degrees with the negative x-axis, and the plane vector will make an acute angle of 50 degrees with the positive x-axis.

Good. Now, from the picture, what are the signs of the x and y components of the 500 km/h vector?
What are the signs of the x and y components of the 150 km/h vector?

 

[physics4ever25]

Good. Now, from the picture, what are the signs of the x and y components of the 500 km/h vector?
What are the signs of the x and y components of the 150 km/h vector?

For the wind vector, the x component would have a negative (-) sign and the y component would have a positive (+) sign. For the plane vector, the x-component would have a positive (+) sign and the y component would have a positive (+) sign.

 

[Chestermiller]

For the wind vector, the x component would have a negative (-) sign and the y component would have a positive (+) sign. For the plane vector, the x-component would have a positive (+) sign and the y component would have a positive (+) sign.

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?