(a) First, convert the truck's velocity from km/h to m/s:
[tex]89.9\,\dfrac{\mathrm{km}}{\mathrm h}\cdot\dfrac{10^3\,\mathrm m}{\mathrm{km}}\cdot\dfrac{\mathrm h}{3600\,\mathrm s}=25.0\,\dfrac{\mathrm m}{\mathrm s}[/tex]
Then the velocity [tex]v[/tex] after traveling [tex]\Delta x=19.3\,\mathrm m[/tex] satisfies
[tex]v^2-{v_0}^2=2a\Delta x[/tex]
where [tex]v_0[/tex] and [tex]a[/tex] are the truck's initial velocity and acceleration, respectively. So we have
[tex]v^2-\left(25.0\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-5.99\,\dfrac{\mathrm m}{\mathrm s^2}\right)(19.3\,\mathrm m)[/tex]
[tex]\implies v^2=394\,\dfrac{\mathrm m^2}{\mathrm s^2}[/tex]
[tex]\implies v=\pm19.8\,\dfrac{\mathrm m}{\mathrm s}[/tex]
Taking the square root gives two possible velocities of the same magnitude but in opposite directions. It's technically feasible that over the given displacement, the truck changes its direction and is accelerating back towards its starting position since we're treating the truck as a particle.
(b) When acceleration is constant, average velocity is given by
[tex]v_{\mathrm{av}}=\dfrac{x-x_0}t=\dfrac{v+v_0}2[/tex]
which tells us that
[tex]t=\dfrac{2(19.3\,\mathrm m)}{v+25.0\,\frac{\mathrm m}{\mathrm s}}[/tex]
Plugging in both possible values of [tex]v[/tex] we found earlier, we get [tex]t=0.862\,\mathrm s[/tex] (if [tex]v[/tex] is positive) or [tex]t=7.42\,\mathrm s[/tex] (if [tex]v[/tex] is negative).
