Solving for the Direction of an Airplane in a Windy Flight

[eshia5857]

Homework Statement

An airplane, whose air speed is 580 km/h, is supposed to fly in a straight path 38 degrees N of E. But a steady 72 km/h wind is blowing from the North. In what direction should the plane head?

Relevant Equations
Pythagorean theorem
Law of sines

Attempt at a solution
The first thing I did was to draw myself a diagram so that I could see what is happening in this problem. I tried to attach a diagram that I made myself, but that did not work out at all.
What I did was to label each one Velocity of the plane with respect to the ground was pointing in the N of E direction and was approximated to 38 degrees, Velocity of the air with respect to the ground was place so that it was head to tail of the previous vector and pointed in the southward direction and the last one was labeled velocity of the plane with respect to the air and it was placed facing in the Eastern direction located directly onto the x-axis. Once I resolve it into components I am getting Y: $$V_{pg}$$*sin($$\theta$$)-$$V_{ag}$$*sin($$\theta$$), for the x component I found $$V_{pg}$$*cos($$\theta$$).I plugged my numbers into the equations and found my y components to have a value of 312.76 km/h and my x components to have a value of 457.05 km/h. I took the arcsin(y/x) and found the direction to be 43.2 degrees this troubles me because the solution is supposed to be 43.6 degrees. I need to know how to fix this error so that I don't make mistakes like this in the future I would appreciate any assistance that can be offered. Thank you for your time.

 

[Sorry!]

Homework Statement

An airplane, whose air speed is 580 km/h, is supposed to fly in a straight path 38 degrees N of E. But a steady 72 km/h wind is blowing from the North. In what direction should the plane head?

Relevant Equations
Pythagorean theorem
Law of sines

Attempt at a solution
The first thing I did was to draw myself a diagram so that I could see what is happening in this problem. I tried to attach a diagram that I made myself, but that did not work out at all.
What I did was to label each one Velocity of the plane with respect to the ground was pointing in the N of E direction and was approximated to 38 degrees, Velocity of the air with respect to the ground was place so that it was head to tail of the previous vector and pointed in the southward direction and the last one was labeled velocity of the plane with respect to the air and it was placed facing in the Eastern direction located directly onto the x-axis. Once I resolve it into components I am getting Y: $$V_{pg}$$*sin($$\theta$$)-$$V_{ag}$$*sin($$\theta$$), for the x component I found $$V_{pg}$$*cos($$\theta$$).I plugged my numbers into the equations and found my y components to have a value of 312.76 km/h and my x components to have a value of 457.05 km/h. I took the arcsin(y/x) and found the direction to be 43.2 degrees this troubles me because the solution is supposed to be 43.6 degrees. I need to know how to fix this error so that I don't make mistakes like this in the future I would appreciate any assistance that can be offered. Thank you for your time.

Was this a textbook question and was it the answer found in the back of the book?
And did you use exact values or rounded values... always use exact values use the calculators memory or store function.

 

[Andrew Mason]

You have to show us how you worked it out. It strikes me that this is a rather difficult vector problem. Unless you show us exactly what you did, it is difficult to find out where you went wrong. What did you find to be the magnitude of the plane to ground velocity? How did you determine that?

AM

 

[eshia5857]

Alright, sorry I was finally able to get an image onto this site, and built up the courage to let you see how wrong I am. I am certain that I am doing something wrong and I have looked at this problem so many times that I thought my eyes would cross. I have to know what is wrong with my logic. Please help anyone, I do not want a solution there is one in the back of my text, I want to understand. I also forgot to include the information Adrew Mason asked about. I have found on few pieces of paper that have now become garbage Vpa, I changed my diagram, used conversions then not using conversions and at some point I know I began to make up some rules of my own out of desperation. I am lost!

 

[eshia5857]

You have to show us how you worked it out. It strikes me that this is a rather difficult vector problem. Unless you show us exactly what you did, it is difficult to find out where you went wrong. What did you find to be the magnitude of the plane to ground velocity? How did you determine that?

AM

Thank you for the last part of your comment it got me really thinking about it again and now I have something more concrete to try, fingers crossed.

 

[Andrew Mason]

Alright, sorry I was finally able to get an image onto this site, and built up the courage to let you see how wrong I am. I am certain that I am doing something wrong and I have looked at this problem so many times that I thought my eyes would cross. I have to know what is wrong with my logic. Please help anyone, I do not want a solution there is one in the back of my text, I want to understand. I also forgot to include the information Adrew Mason asked about. I have found on few pieces of paper that have now become garbage Vpa, I changed my diagram, used conversions then not using conversions and at some point I know I began to make up some rules of my own out of desperation. I am lost!

What you have to do is calculate the length of the plane-ground vector. Call it L. Call the wind velocity vector (length of the air-ground velocity vector) W and the plane-air velocity vector P.

Write out the expression for L in terms of W and P and the known angle. You will get a quadratic equation in terms of L. Find the solution of that equation and then you can determine the angle of P. The correct answer is 43.6 degrees north of east.

AM

 

[emyt]

What you have to do is calculate the length of the plane-ground vector. Call it L. Call the wind velocity vector (length of the air-ground velocity vector) W and the plane-air velocity vector P.

Write out the expression for L in terms of W and P and the known angle. You will get a quadratic equation in terms of L. Find the solution of that equation and then you can determine the angle of P. The correct answer is 43.6 degrees north of east.

AM

Hey, I was trying this as well, I was onto something but I gave up.. could you show your work please?

thanks

 

[Andrew Mason]

Hey, I was trying this as well, I was onto something but I gave up.. could you show your work please?

thanks

You have a vector of indeterminate length going in a direction of 38 degrees north of east (plane-ground velocity - call it L). You have a vector of length 72 (km/hr) from the tail of the previous vector heading due south (W). The difference is the plane-air velocity vector (P)

W + P = L

Work out the length of P in terms of W and L:

P^2 = (W + Lsin38))^2 + (Lcos38)^2

Work that out to find L. You have to solve the quadratic equation (Hint: what is cos^2 + sin^2)?

AM

 

[eshia5857]

There is an attachement in one of the previous parts of this post but I have figured out how to solve it now, and an putting the finishing touches on it, so that it is accurate in everyway. Also I was really confused when I started the problem, take another look at your diagram. My original diagram was causing me to solve the problem wrong. I don't know about just giving out an answer like that, they made me work for mine, now you work for yours. Nothing personal it will just make you better. If you still can't get it in a while, then I will attach my new diagram.
Sorry that message is for emyt, thank you guys very much I appreciate being able to come to a forum where people don't just blurt out answers. Thanks for all the insight, have a great week-end.

 

[emyt]

You have a vector of indeterminate length going in a direction of 38 degrees north of east (plane-ground velocity - call it L). You have a vector of length 72 (km/hr) from the tail of the previous vector heading due south (W). The difference is the plane-air velocity vector (P)

W + P = L

Work out the length of P in terms of W and L:

P^2 = (W + Lsin38))^2 + (Lcos38)^2

Work that out to find L. You have to solve the quadratic equation (Hint: what is cos^2 + sin^2)?

AM

Ahhh I see, I have a diagram like this drawn, I just didn't make the connection.. thanks