Since we are dealing with constant acceleration, we can apply the equations of motion.
Given
Vo = Initial velocity
a = accelaration
t = time
Vo = 1.6 m/s
a = 0.33 m/s2
t = 3.6 s
Procedure
[tex]\begin{gathered} V_f_{}=v_o+a\cdot t \\ V_f=1.6m/s+0.33m/s^2\cdot3.6 \\ V_f=2.79\text{ m/s} \end{gathered}[/tex]
Now, for the distance:
[tex]\begin{gathered} x=v_ot+\frac{1}{2}at^2_{} \\ x=1.6\text{ m/s}\cdot3.6\text{ s+}\frac{1}{2}\cdot0.33\cdot3.6^2 \\ x=7.9\text{ m} \end{gathered}[/tex]