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Help wanted

by physicsss
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physicsss
#1
Sep13-04, 11:25 PM
P: 319
A light plane is headed due south with a speed relative to still air of 200 km/h. After 1 hour, the pilot notices that they have covered only 190 km and their direction is not south but southeast. What is the wind velocity in km/h and what angle it it north of east?
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Tide
#2
Sep13-04, 11:29 PM
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The velocity of the wind is the difference between the actual velocity of the plane and the "intended" velocity of the plane.
physicsss
#3
Sep13-04, 11:49 PM
P: 319
not really...in the book the intended is 240 km/h but the actual is 180 km and the answer they give is 170 k/h with an angle of 41.5 N of E.

Tide
#4
Sep14-04, 12:02 AM
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Help wanted

Quote Quote by physicsss
not really...in the book the intended is 240 km/h but the actual is 180 km and the answer they give is 170 k/h with an angle of 41.5 N of E.
Remember, velocity is a vector!
physicsss
#5
Sep14-04, 12:26 AM
P: 319
but there's no angles given.
Tide
#6
Sep14-04, 12:34 AM
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Due south and southeast sound like angles to me! :-)
physicsss
#7
Sep14-04, 02:16 AM
P: 319
i still dont get it. angle is 45 degrees?
Tide
#8
Sep14-04, 02:31 AM
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LaTex doesn't seem to be working so I'll try this:

The actual speed of the plane is 190/sqrt(2) km/h so it's actual velocity is (1, -1)*190/sqrt(2).

The "intended velocity of the plane is (0, -1)*200 km/h.

The difference between them is the wind velocity:
(190/sqrt(2), 200 - 190/sqrt(2))

Now just divide the y component by the x component to find the tangent of the angle you're looking for.

A drawing might help you see it better.
HallsofIvy
#9
Sep14-04, 08:17 AM
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Thanks
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Quote Quote by Tide
A drawing might help you see it better.
That should be the first thing you do!
physicsss
#10
Sep14-04, 12:17 PM
P: 319
Tide, Does Latex work now? I still don't get the answers that the book got using what you did...


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