Use Newton's second law to determine the acceleration applied by the stopping force:
[tex]F=ma\implies7024\,\mathrm N=(900\,\mathrm{kg})a\implies a\approx7.8\,\frac{\rm m}{\mathrm s^2}[/tex]
Then recall that
[tex]{v_f}^2-{v_i}^2=2a\Delta x[/tex]
Take the direction in which the car is traveling to be positive, so that its acceleration points in the opposite direction and [tex]a[/tex] has negative sign. Then
[tex]\left(2\,\frac{\rm m}{\rm s}\right)^2-\left(18\,\frac{\rm m}{\rm s}\right)^2=2\left(-7.8\,\frac{\rm m}{\mathrm s^2}\right)\Delta x\implies\Delta x\approx21\,\mathrm m[/tex]
