#### Answer

The boat is moving with a speed of 9.2 m/s at an angle of $19.6^{\circ}$ north of east.

#### Work Step by Step

We can find the eastward component of velocity after the 6.0 second acceleration period.
$v_x = v_{0x}+a_xt$
$v_x = 5.0~m/s+(0.80~m/s^2)~cos(40^{\circ})(6.0~s)$
$v_x = 8.68~m/s$
We can find the northward component of velocity after the 6.0 second acceleration period.
$v_y = v_{0y}+a_yt$
$v_y = 0+(0.80~m/s^2)~sin(40^{\circ})(6.0~s)$
$v_y = 3.09~m/s$
We can find the boat's speed.
$v = \sqrt{(v_x)^2+(v_y)^2}$
$v = \sqrt{(8.68~m/s)^2+(3.09~m/s)^2}$
$v = 9.2~m/s$
We can find the angle north of east.
$tan(\theta) = \frac{v_y}{v_x}$
$\theta = arctan(\frac{3.09~m/s}{8.68~m/s})$
$\theta = 19.6^{\circ}$
The boat is moving with a speed of 9.2 m/s at an angle of $19.6^{\circ}$ north of east.