Finding the velocity of the wind

[xChee]

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

V_g = V_a + V_w

The Attempt at a Solution

so i know we're suppose to use the formula V_g = V_a + V_w

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

 

[berkeman]

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

V_g = V_a + V_w

The Attempt at a Solution

so i know we're suppose to use the formula V_g = V_a + V_w

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?

 

[xChee]

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

 

[berkeman]

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.

 

[AmritpalS]

I worked it out. It is, indeed, 102.425km/h -13° from the positive x axis, or using your notation South 76.69° East.