# Thread: Application - Airplane and Wind

1. ## Application - Airplane and Wind

the course and ground speed of a plare are 70 degree and 400 miles per hour respectively. There is a 60 mph wind blowing from the south. find the approximate direction and air speed of the plane.

This is how I learned how to solve it with vectors:

$\displaystyle v_1 = 400 \langle \cos {70}, \sin {70} \rangle$
$\displaystyle v_1 = \langle 253.33, 309.56 \rangle$

Would the vector for the wind be:

$\displaystyle v_2 = 60 \langle \cos {270}, \sin {270} \rangle$ ?

I'm confused on the part of the question that says the wind is blowing from the south. Help?

Then you would just add the two vectors and find the resultant for the speed of the airplane right? How would you find the direction of the airplane?

2. Originally Posted by chrozer
the course and ground speed of a plare are 70 degree and 400 miles per hour respectively. There is a 60 mph wind blowing from the south. find the approximate direction and air speed of the plane.

This is how I learned how to solve it with vectors:

$\displaystyle v_1 = 400 \langle \cos {70}, \sin {70} \rangle$
$\displaystyle v_1 = \langle 253.33, 309.56 \rangle$

Would the vector for the wind be:

$\displaystyle v_2 = 60 \langle \cos {270}, \sin {270} \rangle$ ?

I'm confused on the part of the question that says the wind is blowing from the south. Help?

Then you would just add the two vectors and find the resultant for the speed of the airplane right? How would you find the direction of the airplane?
assuming the author of the problem is using directions relative to the +x axis ... (and that is a HUGE assumption)

$\displaystyle A_x + W_x = G_x$

$\displaystyle A_x + 0 = 400\cos(70)$

$\displaystyle A_x = 400\cos(70)$

$\displaystyle A_y + W_y = G_y$

$\displaystyle A_y + 60 = 400\sin(70)$

$\displaystyle A_y = 400\sin(70) - 60$

$\displaystyle |A| = \sqrt{A_x^2 + A_y^2} \, \approx 344$ mph

direction, $\displaystyle \theta = \arctan\left(\frac{A_y}{A_x}\right) \approx 67^{\circ}$

3. Originally Posted by skeeter
assuming the author of the problem is using directions relative to the +x axis ... (and that is a HUGE assumption)

$\displaystyle A_x + W_x = G_x$

$\displaystyle A_x + 0 = 400\cos(70)$

$\displaystyle A_x = 400\cos(70)$

$\displaystyle A_y + W_y = G_y$

$\displaystyle A_y + 60 = 400\sin(70)$

$\displaystyle A_y = 400\sin(70) - 60$

$\displaystyle |A| = \sqrt{A_x^2 + A_y^2} \, \approx 344$ mph

direction, $\displaystyle \theta = \arctan\left(\frac{A_y}{A_x}\right) \approx 67^{\circ}$
Yeah I had to assume that as I had always been thought to do it like that. I see now. Thanks alot.