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 June 12th, 2019, 05:15 PM #1 Senior Member   Joined: Mar 2019 From: TTF area Posts: 136 Thanks: 1 Plane A plane flys due east at a constant air speed of 151ms^-1. A wind blows at a constant speed of 64ms^-1 from the north. Calculate the resultant speed and the direction of the aircraft. So I was thinking. V^2=151^2+64^2 . V=164km h^-1 Tan θ= 64/151 θ=23degrees The air speed is 164km in the direction 23degrees east. ?
June 12th, 2019, 05:34 PM   #2
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Quote:
 Originally Posted by helpmeddddd A plane flys due east at a constant air speed of 151ms^-1. A wind blows at a constant speed of 64ms^-1 from the north. Calculate the resultant speed and the direction of the aircraft. So I was thinking. V^2=151^2+64^2 . V=164km h^-1 Tan θ= 64/151 θ=23degrees The air speed is 164km in the direction 23degrees east. ?
Almost. The magnitude and the angle have the correct numerical values, but such a plane can't be flying due east. You need to say that it's 23 degrees south of east (or 23 degrees S of E.)

-Dan

 June 12th, 2019, 05:54 PM #3 Senior Member   Joined: Mar 2019 From: TTF area Posts: 136 Thanks: 1 Thanks. For practice purposes, let's say the aircraft came somewhere from California and went to Kansas. On the return journey, the pilot wants to fly directly back from Kansas to California . The air speed of the plane is 151ms^-1 and the wind still blows at a constant speed of 64ms^-1 from the north. In what direction must the plane head to reach California? West? What will be the resultant speed? Would this just be the same? Last edited by skipjack; June 13th, 2019 at 01:45 AM.
June 12th, 2019, 06:09 PM   #4
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Quote:
 Originally Posted by helpmeddddd Thanks. For practice purposes, let's say the aircraft came somewhere from California and went to Kansas. On the return journey, the pilot wants to fly directly back from Kansas to California . The air speed of the plane is 151ms^-1 and the wind still blows at a constant speed of 64ms^-1 from the north. In what direction must the plane head to reach California? West? What will be the resultant speed? Would this just be the same?
The same speed. Since we now want to fly in a straight line we would need to angle the plane to correct for the wind. So we'd fly 23 degrees N of W.

-Dan

Last edited by skipjack; June 13th, 2019 at 01:45 AM.

 June 13th, 2019, 12:41 PM #5 Senior Member   Joined: Mar 2019 From: TTF area Posts: 136 Thanks: 1 If the plane now is going N of W won't the resultant speed be different ? since the constant speed of the wind is going 64ms^-1 north
June 13th, 2019, 02:17 PM   #6
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Quote:
 Originally Posted by helpmeddddd If the plane now is going N of W won't the resultant speed be different ? since the constant speed of the wind is going 64ms^-1 north
The Pythagorean theorem does not worry about "direction" in terms of positives and negatives. (The speeds get squared so the negative drops out.)

The wind is coming from the North, not directed to the North. So the pilot has to angle the plane toward the North a bit to compensate.

-Dan

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