The car's velocity at time [tex]t[/tex] is
[tex]v(t)=16\dfrac{\rm m}{\rm s}+\displaystyle\int_0^t\left(\left(-0.50\frac{\rm m}{\mathrm s^2}\right)u\right)\,\mathrm du=16\dfrac{\rm m}{\rm s}+\left(-0.25\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]
It comes to rest at
[tex]v(t)=0\implies16\dfrac{\rm m}{\rm s}=\left(0.25\dfrac{\rm m}{\mathrm s^2}\right)t^2\implies t=8.0\,\mathrm s[/tex]
Its velocity over this period is positive, so that the total distance the car travels is
[tex]\displaystyle\int_0^{8.0}v(t)\,\mathrm dt=\left(16\dfrac{\rm m}{\rm s}\right)(8.0\,\mathrm s)+\frac13\left(-0.25\dfrac{\rm m}{\mathrm s^2}\right)(8.0\,\mathrm s)^3=\boxed{85\,\mathrm m}[/tex]
so the answer is D.
