Two bodies are moving such that the velocity of one is twice that of the other and they make an angle of `60^@` with each other. Find the relative velocity of one with respect to the other.

Explanation

Representing the velocities in the vector diagram [Fig.2.42], we get ,`vec(AB)=vecu`= velocity of one body , `vec(AC)=2vecu`=velocity of the other.

The relative velocity of the second with respect to the first,
`vecw=vec(AC)-vec(AB)=vec(AC)+vec(BC)` [from triangle law of vector addition]
From `Delta ABC`
`w^2=(-u)^2+(2u)^2+2*(-u)*2u cos 60^@`
`=5u^2-2u^2=3u^2`
`therefore w=sqrt(3)u`
Also from the trigonometric rule,
`(2u)/(sin angleABC)=(w)/(sin 60^@)=(sqrt(3)xx2)/(sqrt(3))`
`or, sin angle ABC=1 sin 90^@`
`therefore angleABC=90^@`.
Hence,`vecw,i.e., vec(BC)` is perpendicular to `vecu, i.e., vec(AB)`.
Similarly, the relative velocity of the first with respect to the second can be found . In this case , the relative velocity will be `-vecw` as the direction will be opposite.