The velocituy of particle (P) due east is ` 4 ms^(-1)` due South,. What is the velocity of ` P w.r.t. Q` ?

Explanation

In Fig. 2 (c ) . 90. ` vec v_P = vec(OA)=4 m//s` due East ,
` vec v_Q = vec (OB) = 3 m//s ` due South
.
` vec _(PQ) = vec _P - vec _Q = vec _P + (- vec v_Q) = vec (OA) + vec (OD)`
:. ` | vec v_(PQ) | = sqrt ( 4^@ + 3^2) = 5 ms^(-1)`,
` tan theta = (AV)/(OA) = (OD)/(OA) = 3/= 0. 75`
or ` therta = 36^@ 52' ` North of East .