If x is the distance of point P from O then from Fig. 5.63
`tan phi=(x//h)`
or `x=h tan phi`
or `(dx)/(dt)=h(sec^(2)phi)(dphi)/(dt)`
i.e., `v=h"sec"^(2)phiomega` [as `(dx//dt)`=v and `(dphi//dt)=omega`]
Here, h=3m, `phi=180^(@)-(45^(@)+90^(@))=45^(@)` and `omega=0.1` rad/s
So, `v=3xx(sqrt(2))^(2)xx0.1=`0.6m/s