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Old Sep 19th 2008, 09:37 AM   #1
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Post jet airliner

A jet airliner moving at 500 mph due east moves into a region where the wind is blowing at 120 mph in a direction 30.0 north of east. What is the new velocity and direction of the aircraft?

Last edited by tweedydaf; Sep 19th 2008 at 02:25 PM.
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Old Sep 19th 2008, 07:45 PM   #2
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Originally Posted by tweedydaf View Post
A jet airliner moving at 500 mph due east moves into a region where the wind is blowing at 120 mph in a direction 30.0 north of east. What is the new velocity and direction of the aircraft?
Resolve the x- y- component into 120cos 30 and 120sin 30 respectively
resultant vector magnitude
= sqrt[(120cos30+500)^2+(120sin30)^2]
=607mph, cor to 3 sig fig

the direction is N 5.67 E which equals arctan[(120sin30)/ (120cos30+500)]
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Old Sep 19th 2008, 07:46 PM   #3
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Originally Posted by tweedydaf View Post
A jet airliner moving at 500 mph due east moves into a region where the wind is blowing at 120 mph in a direction 30.0 north of east. What is the new velocity and direction of the aircraft?
The plane will fly with a velocity that is the vector sum of the velocity of the wind and the velocity the jet can fly in still air. So we are looking for the vector sum of 500 mph due E and 120 mph 30 degrees N of E.

I'd choose the "usual" coordinate system: +x in the direction of E and +y in the direction of N. To add these vectors we need to break them down into components in these coordinate directions. Let's start with the wind, which I will call vector A:
$\displaystyle A_x = 120~cos(30^o) = 103.923$

$\displaystyle A_y = 120~sin(30^o) = 60$

Now for the jet, which I will call vector B:
$\displaystyle B_x = 500$

$\displaystyle B_y = 0$
(since the plane is trying to fly due E there is no angle. Or if you like, the angle is 0 degrees.)

Now we add the like components to form our final velocity vector R:
$\displaystyle R_x = A_x + B_x = 603.923$

$\displaystyle R_y = A_y + B_y = 60$

To find the magnitude of this vector we use the Pythagorean Theorem:
$\displaystyle R = \sqrt{R_x^2 + R_y^2} = 606.896~mph$

To find the direction we note that the vector R is in the second quadrant. Then
$\displaystyle tan(\theta) = abs \left ( \frac{R_y}{R_x} \right ) \implies \theta = 5.6461^o$

So your final velocity of the plane will be 607 mph at 5.65 degrees N of W.

-Dan
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