One more Vector problem and we are good to go

[junior_J]

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 ?

 

[radou]

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"

 

[junior_J]

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 !

 

[junior_J]

err ... got it ! I am so embarrased !

thank you ! :)

 

[radou]

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.