Answer
$v_{2g}=1.72m/s$
$23.8^{\circ}$ south of west.
Work Step by Step
We know that
$\vec {v_{1g}}=\vec v_{12}+\vec v_{2g}$
Where $\vec {v_{1g}},\vec v_{12}$ and $\space \vec v_{2g}$ represent velocity of boat 1 relative to ground, velocity of boat 1 relative to boat 2, and velocity of boat 2 relative to ground.
The above equation can be rearranged as:
$\vec {v_{2g}}=\vec {v_{1g}}-\vec {v_{12}}$
We plug in the known values to obtain:
$\vec{v_{2g}}=(0.775m/s)\hat y-[(2.15m/s)sin 47.0^{\circ}\hat x+(2.15m/s)cos 47.0^{\circ}\hat y]$
$\vec {v_{2g}}=(-1.57m/s)\hat x+(-0.691m/s)\hat y$
Now the magnitude of $\vec{v_{2g}}$ is given as
$v_{2g}=\sqrt{v_{2g,x}^2+v_{2g,y}^2}$
We plug in the known values to obtain:
$v_{2g}=\sqrt{(-1.57)^2+(0.691)^2}=1.72m/s$
and the direction of $\vec{v_{2g}}$ is given as
$\theta=tan^{-1}(\frac{v_{2g,y}}{v_{2g,x}})$
$\theta=tan^{-1}(\frac{-0.691}{-1.57})=23.8^{\circ}$ south of west.
