A car is moving at `80 km*h^(-1)` towards north. Another car is moving at `80 sqrt(2) km*h^(-1)` towards north-west. Find the relative velocity of the second car with respect of the first.

Explanation

In [Fig.2.41], the velocity of the first car, `vecu =80 km*h^(-1)` towards north `=vec(AB)`, the velocity of the 2nd car, `vecv=80sqrt(2)km*h^(-1)` towards north -west =`vec(AC)`.

Hence, relative velocity of the second car with respect to teh first,
`vecw=vecv-vecu=vecv+(vec(-u))=vec(AC)+vec(BA)=vec(BC)` [from triangle law of vector addition]
As `angle BAC=45^@`, we get from `DeltaABC`,
`w^2=u^2+v^2-2uvcos 45^@`
`=(80)^2+(80sqrt(2)^2)-2.80*80sqrt(2)*1/sqrt(2)`
`w^2=80^2[1+(sqrt(2)^2)-2]=80^2`
`therefore w=80km*h^(-1)`
Since, w=u, from `Delta ABC`,
`angle ACB=angleBAC=45^@ therefore angleABC=90^@`
Hence, the vector `vec(BC) =vecw` is directed towards west.