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Finding the velocity of the wind

 
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May21-12, 03:51 PM   #1
 

Finding the velocity of the wind


1. The problem statement, all variables and given/known data

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

2. Relevant equations

Vg = Va + Vw

3. 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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May21-12, 05:57 PM   #2
 
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Quote by xChee View Post
1. The problem statement, all variables and given/known data

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

2. Relevant equations

Vg = Va + Vw

3. 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?
 
May21-12, 06:28 PM   #3
 
Quote by berkeman View Post
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
 
May21-12, 06:32 PM   #4
 
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Finding the velocity of the wind


Quote by xChee View Post
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.
 
May21-12, 08:42 PM   #5
 
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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