Recall that
[tex]{v_f}^2-{v_i}^2=2a\Delta x[/tex]
where [tex]v_i[/tex], [tex]v_f[/tex] are the truck's initial and final velocities, respectively; [tex]a[/tex] is its acceleration; and [tex]\Delta x[/tex] is the total change in its position. We have everything we need in order to solve for [tex]v_f[/tex]. Note that the truck's acceleration is always positive, so we should expect the final velocity to also be positive.
[tex]{v_f}^2-\left(3\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(1.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(50\,\mathrm m)[/tex]
[tex]\implies{v_f}^2=129\,\dfrac{\mathrm m^2}{\mathrm s^2}[/tex]
[tex]\implies v_f=11.4\,\dfrac{\mathrm m}{\mathrm s}[/tex]
