Answer
$41.7\space m$
Work Step by Step
Please see the attached image first.
First of all, let's find the acceleration vector $\vec a$. Here we assume the initial direction of the sailboard in $x+$ direction.
we can write,
$a_{x}=0.47\space m/s^{2}$
$a_{y}=0.54\space m/s^{2}$
Therefore, $\vec a= (0.47i+0.54j)\space m/s^{2}$
Now we apply equation $S=ut+\frac{1}{2}at^{2}$ separately to x,y directions to find the position vector $\vec S$ with respect to the initial position.
$\rightarrow S=ut+\frac{1}{2}at^{2}$
Let's plug known values into this equation.
$S_{x}= 6.28\space m/s\times5.42\space s+ \frac{1}{2}\times(0.47\space m/s^{2})(5.42\space s)^{2}$
$S_{x}=34.04\space m+6.9\space m$
$S_{x}= 40.94\space m$
$\uparrow S=ut+\frac{1}{2}at^{2}$
Let's plug known values into this equation.
$S_{y}=0+\frac{1}{2}\times(0.54\space m/s^{2})(5.42\space s)^{2}$
$S_{y}=7.93\space m$
$\vec S= (40.94i+7.93j)\space m$
The magnitude of boards displacement $= \sqrt {S_{x}^{2}+S_{y}^{2}}$
$$=\sqrt {(40.94\space m)^{2}+(7.93\space m)^{2}}=41.7\space m$$
