Vector Word Problem

[e^x]

I've set up two vectors, AB pointing north with a magnitude of 90, and AC pointing northeast with a magnitude of 80. I know the angle between them is 5 degrees, but I don't know what to do next.

 

[lev888]

View attachment 29598
I've set up two vectors, AB pointing north with a magnitude of 90, and AC pointing northeast with a magnitude of 80. I know the angle between them is 5 degrees, but I don't know what to do next.

You have 3 vectors in play:
Plane's velocity, wind, actual direction and distance of plane's travel. What's the relationship between them?

 

[e^x]

You have 3 vectors in play:
Plane's velocity, wind, actual direction and distance of plane's travel. What's the relationship between them?

The actual direction and distance has been changed by the wind? Idk what to do though. I can't set up a right triangle with the two vectors, and I don't think dot product or cross product would be helpful either

 

[blamocur]

The actual direction and distance has been changed by the wind? Idk what to do though. I can't set up a right triangle with the two vectors, and I don't think dot product or cross product would be helpful either

I think you are going in the right direction. I would express AB and AC in cartesian coordinates, .e.g Y pointing north and X pointing east. Then I would rephrase lev888's question: what is the relation between speed vectors of plane's air speed, plane's ground speed and the wind.

 

[topsquark]

Let $\textbf{v}_{pg}$ be the plane's actual velocity (with respect to the ground), $\textbf{v}_{pa}$ be the plane's speed in still air, and $\textbf{v}_{ag}$ be the speed of the air with respect to the ground. Then
$\textbf{v}_{pg} = \textbf{v}_{pa} + \textbf{v}_{ag}$

Can you continue?

-Dan

 

[e^x]

Let $\textbf{v}_{pg}$ be the plane's actual velocity (with respect to the ground), $\textbf{v}_{pa}$ be the plane's speed in still air, and $\textbf{v}_{ag}$ be the speed of the air with respect to the ground. Then
$\textbf{v}_{pg} = \textbf{v}_{pa} + \textbf{v}_{ag}$

Can you continue?

-Dan

So the velocity is the actual velocity minus the speed in still air? I don't think it's that simple...

 

[lev888]

So the velocity is the actual velocity minus the speed in still air? I don't think it's that simple...

See post 5.

 

[e^x]

I think you are going in the right direction. I would express AB and AC in cartesian coordinates, .e.g Y pointing north and X pointing east. Then I would rephrase lev888's question: what is the relation between speed vectors of plane's air speed, plane's ground speed and the wind.

Okay, so with 80sin5 and 80cos5 found the components of the plane's velocity in the x and y directions to be 6.972 and 79.696, respectively. I then subtracted this vector from the vector <0i ,90j> and got <-6.972i, -10.304j>. Is this correct? And where do I start if I want to solve b, in which direction should the pilot have flown in order to get to their desired destination?

 

[blamocur]

Looks good, except that one of the signs in your result is reversed. Worth remember that the numbers you got show how much wind travels in 30 min, not in one hour. Once you get the wind speed coordinates you need the speed itself, a.k. the vector's length.

When you done with the wind speed you can right an equation (either in vector notation or in coordinates) for the plane's air speed needed to achieve ground speed vector of 180 km/h north.

 

[e^x]

Looks good, except that one of the signs in your result is reversed. Worth remember that the numbers you got show how much wind travels in 30 min, not in one hour. Once you get the wind speed coordinates you need the speed itself, a.k. the vector's length.

When you done with the wind speed you can right an equation (either in vector notation or in coordinates) for the plane's air speed needed to achieve ground speed vector of 180 km/h north.

Okay, so I found the vector's length and doubled it to get a wind speed of 24.882 km/hr. But then how do I find what angle the pilot should have gone in?

 

[blamocur]

Okay, so I found the vector's length and doubled it to get a wind speed of 24.882 km/hr. But then how do I find what angle the pilot should have gone in?

I get the same number, so it must be right
Now you know by which vector the wind shifted the plane off the course. What remains is to adjust the plane's course so that the same wind vector shifts it to the right course.

 

[e^x]

I get the same number, so it must be right
Now you know by which vector the wind shifted the plane off the course. What remains is to adjust the plane's course so that the same wind vector shifts it to the ri

So I have to find the angle, right? I thought a dot product could yield the angle between the desired vector and the wind velocity, but this brought me to an inverse cosine of 7.748, which doesn't exist.

 

[skeeter]

air vector + wind vector = track vector

$(0,90) + \dfrac{1}{2}(w_x,w_y) = (80\cos(85),80\sin(85))$

wind speed = $\sqrt{w_x^2 + w_y^2}$

wind direction, $\theta = \arctan\left(\dfrac{w_y}{w_x}\right)$

 

[blamocur]

So I have to find the angle, right? I thought a dot product could yield the angle between the desired vector and the wind velocity, but this brought me to an inverse cosine of 7.748, which doesn't exist.

Not sure where you got inverse cosine (a.k.a arccosine or arccos) of 7.748 from. To figure out the angle between two vectors you need arccos of a dot product of unit vectors, not arbitrary length ones. Moreover, unless the wind is orthogonal to the plane course just adjusting the angle by the same amount (i.e. angle between AB and AC) will not give you the solution, but only an approximation.

I would solve this in vector form. You've already introduced vectors AB for the desired course and AC for the actual one because of the wind vector BC. Now find a vector, say AD, such that the wind will shift it to AC. Once you done that the angle between AD and AC is your answer.

Good luck.