One more Vector problem and we are good to go

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The problem involves calculating the change in velocity of a yacht transitioning from 5 m/s due north to 3 m/s due west. The calculated resultant velocity is 5.8 m/s, and the direction is initially found to be 31 degrees west of north. However, the correct interpretation shows that the change in velocity points south-west, leading to the conclusion that the direction is actually 31 degrees west of south. The confusion arises from the orientation of the vectors, where north is up and south is down on the diagram. Understanding vector subtraction clarifies the direction of the resultant velocity.
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One more Vector problem and we are good to go!

The Problem:
Find the change in velocity of a yacht if it changes its velocity from
5ms^-1 due north to 3ms^-1 due west and also state its direction.

The solution :
(resultant vel.)^2 = 5^2 + 3^2 = 5.8 ms^-1
direction = tan-1(3/5) = 31 degrees west of north

However , the books answer is 5.8 ms^-1 in a direction 31 degrees west of south . what am i doing wrong here ?
 
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Draw down the vectors and it should become clear why the direction is 31 degrees west of south. Btw, the change equals \vec{v}_{2}-\vec{v}_{1}.

http://upload.wikimedia.org/wikipedia/commons/thumb/5/5b/Vector_subtraction.png/217px-Vector_subtraction.png"
 
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I did draw it but unless south means up and norths means down on paper I don't know what ur getting at lolz I want to tear my book apart !
 
err ... got it ! I am so embarrased !

thank you ! :)
 
junior_J said:
I did draw it but unless south means up and norths means down on paper I don't know what ur getting at lolz I want to tear my book apart !

\vec{v}_{1} is pointing north (up). \vec{v}_{2} is pointing west (left). \vec{v}_{2}-\vec{v}_{1} is then pointing south-west.
 

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