A man A moves due to East with velocity `6 ms^(-1)` and another man B moves in `N-30^(@)E` with `6 ms^(-1)` . Find the velocity of B w.r.t. A.

Explanation

Given, `v_(A)=6hat(i)`,

`v_(B)=v_(B) cos 60^(@)hat(i)+v_(B)sin 60^(@)hat(j)`
`=6((1)/(2))hat(i)+6((sqrt(3))/(2))hat(j)=3hat(i)+3sqrt(3)hat(j)`
To find the velocity,
`v_(BA)=v_(B)-v_(A)=(3hat(i)+3sqrt(3)hat(j))-6hat(i)=-3hat(i)+3sqrt(3)hat(j)`
`|v_(BA)|=sqrt((-3)^(2)+(3sqrt(3))^(2))`
`=sqrt(9+27)=sqrt(36)=6 ms^(-1)`
Here , `hat(i)` is -ve and `hat(j)` is +ve. So, second quadrant is possible. Direction,
`tan alpha = ("coefficient of"hat(j))/("coefficient of"hat(i))=(3sqrt(3))/(-3)=-sqrt(3) rArr alpha = 60^(@)`