Answer
4.6 m/s
Work Step by Step
Please see the attached image first.
Let's take,
Velocity of the passenger relative to the shore = $V_{PS}$
Velocity of the passenger relative to the boat 2 = $V_{P2}$
Velocity of the boat 2 relative to the shore = $V_{2S}$
Velocity of boat 2 relative to boat 1 = $V_{21}$
Velocity of the boat 1 relative to the shore = $V_{1S}$
We can write,
$V_{PS}=V_{P2}+V_{2S}-(1)$
$V_{2S}=V_{21}+V_{1S}-(2)$
(2)=>(1),
$V_{PS}=V_{P2}+V_{21}+V_{1S}$ ; So we can draw this in the diagram given below.
Using the Pythagorean theorem we can get,
$V_{PS}=\sqrt {V_{PSx}^{2}+V_{PSy}^{2}}$
$V_{PS}=\sqrt {(V_{P2}+V_{21}cos30^{\circ})^{2}+(V_{21}sin30^{\circ}+V_{1S})^{2}}$
$V_{PS}=\sqrt {(1.2\space m/s+1.6cos30^{\circ}m/s)^{2}+(1.6sin30^{\circ}m/s+3\space m/s)^{2}}$
$V_{PS}=\sqrt {(2.59\space m/s)^{2}+(3.8\space m/s)^{2}}=4.6\space m/s$
