Solution:

To find the resultant velocity using unit vectors:

- Initial velocity: \( \vec{u} = 0.4 \, \hat{i} \, \text{m/s} \)
- Acceleration: \( \vec{a} = 0.15 \, \hat{j} \, \text{m/s}^2 \)
- Time: \( t = 2 \, \text{seconds} \)

Using the equation of motion \( \vec{v} = \vec{u} + \vec{a}t \):

- In the \( x \)-direction: \( v_x = 0.4 \, \text{m/s} \)
- In the \( y \)-direction: \( v_y = 0 + 0.15 \times 2 = 0.3 \, \text{m/s} \)

The resultant velocity is:

\[
|\vec{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{(0.4)^2 + (0.3)^2} = \sqrt{0.25} = 0.5 \, \text{m/s}
\]

Thus, the resultant velocity is 0.5 m/s