Finding the velocity of the wind

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To find the wind velocity affecting an airplane traveling N40°E at 1000 km/h, the ground speed is measured at N45°E and 1050 km/h. Using the equation Vg = Va + Vw, the problem involves vector subtraction. The calculated wind velocity is approximately 102.4 km/h, confirmed through vector analysis and the cosine law. Additionally, converting vectors to rectangular coordinates and back to polar coordinates is suggested for verification. The final wind velocity is approximately 102.425 km/h at an angle of South 76.69° East.
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Homework Statement



An airplane travels N40∘E at an airspeed of 1000km/h. Measurement on the ground shows that the plane is traveling N45∘E at a speed of 1050 km/h. Calculate the velocity of the wind.

Homework Equations



Vg = Va + Vw

The Attempt at a Solution



so i know we're suppose to use the formula Vg = Va + Vw

we know Vg is 1050km/h N45∘E
and Va is 1000km/h N40∘E
we don't know Vw, so we subtract it to the other side of the equation and we have a subtraction of vectors

I drew my vector diagram, and applied to cos law to get the veloctiy of the wind,

a= sqrt[1050^2 + 1000^2 -2(1000)(1050)cos∘]
and got 102.4 km/h as my wind velocity,

can someone confirm if I'm right? :s
 
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xChee said:

Homework Statement



An airplane travels N40∘E at an airspeed of 1000km/h. Measurement on the ground shows that the plane is traveling N45∘E at a speed of 1050 km/h. Calculate the velocity of the wind.

Homework Equations



Vg = Va + Vw

The Attempt at a Solution



so i know we're suppose to use the formula Vg = Va + Vw

we know Vg is 1050km/h N45∘E
and Va is 1000km/h N40∘E
we don't know Vw, so we subtract it to the other side of the equation and we have a subtraction of vectors

I drew my vector diagram, and applied to cos law to get the veloctiy of the wind,

a= sqrt[1050^2 + 1000^2 -2(1000)(1050)cos∘]
and got 102.4 km/h as my wind velocity,

can someone confirm if I'm right? :s

A good way that you can confirm the answer yourself is to use component-wise subtraction, and convert that back into polar coordinates. Are you familiar with converting back and forth between polar and rectangular coordinates?
 
berkeman said:
A good way that you can confirm the answer yourself is to use component-wise subtraction, and convert that back into polar coordinates. Are you familiar with converting back and forth between polar and rectangular coordinates?

no... :l
 
xChee said:
no... :l

Here is a referenc for you then (partway down this page):

http://en.wikipedia.org/wiki/Polar_coordinate_system

Vectors are usually added and subtracted in rectangular coordinates, so a natural way to do your problem is to convert the vectors you are given into their x and y components in rectangular coordinates, do the subtraction, and then convert the answer back into the polar notation of the problem.
 
I worked it out. It is, indeed, 102.425km/h -13° from the positive x axis, or using your notation South 76.69° East.
 

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