Correct Answer - A
The velocity of car A and car B are represented by the vectors in the direction as given in the question as shown in the figure.
Given : `v_(A)=v_(B)=v=5` km/minute
The relative of car B w.r.t to car A is given by
`vecv_(BA)=vecv_(B)-vecV_(A)`
`|vecv_(BA)|=sqrt(v_(B)^(2)+(-vecv_(A))^(2)+2(v_(B))(-v_(A))cos(-90)^(@))`
`=sqrt((5)^(2)+(-5)^(2)+(0))=5sqrt(2) km//"minute"`
The direction `beta` is given by
`tan beta=(v_(B))/(-v_(A))=(5)/(-5)=-1` or `beta=45^(@)` in south east.
