Answer
$16.3\;\rm m/s$
Work Step by Step
To find its final velocity at $t=2$ s we need to use the impulse-momentum theorem.
$$\Delta p_x=J_x$$
where the impulse is given by
$$J_x=\int F_xdt$$
and $$\Delta p_x=m v_{fx}-mv_{ix} $$
Thus,
$$m v_{fx}-mv_{ix} =\int_{-2}^{2} F_xdt$$
Solving for $v_{fx}$
$$ v_{fx} =\dfrac{mv_{ix}+\int_{-2}^{2} F_xdt}{m}= v_{ix}+\dfrac{1}{m}\int_{-2}^{2} F_xdt$$
Plugging $F_x$ formula from the problem's text.
$$ v_{fx} = v_{ix}+\dfrac{1}{m}\int_{-2}^{2} (4-t^2)dt$$
$$ v_{fx} = v_{ix}+\dfrac{1}{m}\left[4t-\dfrac{t^3}{3}\right]_{-2}^2$$
$$ v_{fx} = v_{ix}+\dfrac{1}{m}\left[\left( 4(2)-\dfrac{(2)^3}{3}\right)-\left(4(-2)-\dfrac{(-2)^3}{3}\right)\right] $$
$$ v_{fx} = v_{ix}+\dfrac{32}{3m} $$
Plugging the known;
$$ v_{fx} = -5 +\dfrac{32}{3\times 0.5}=\color{red}{\bf 16.3}\;\rm m/s $$
