Find the instantneous axis of rotation of a rod length `l` when its end `A` moves with a velocity `vecv_(A)=Vi` and the rod rotates with an angular velocity `vecomega=-v/(2l)hatk`.

Explanation

Let us choose the point `P` as `ICR` in the extended rod. We can say `ICR` is a point of zero velocity. So we can write
`vecv_(P)=vecv_(P,A)+vecv_(A)`
we have `vecv_(P)=0`
Hence `vecv_(P,A)=vecv_(A)=0`

Here `vec_(A)=vhati` and `vecv_(P,A)=-omegahati`
Hence `-omegarhati+vhati=0`
`implies v=omegar`
or `r=v/omega=v/((v/2l))=2l`
Hence `ICR` will be located at a distance `2l` from `A`.
Method 2
The instantaneous centre of rotation will be lie on the perpendicular line at point `A` with `vecv_(A)`.
The `ICR` will lie at distance `AP+(|vecv_(A)|)/omega=v/(v/2l)=2l`
or at a distance `2l` from point `A`.