Use the kinematic formula
[tex]{v_2}^2 - {v_1}^2 = 2a\Delta x[/tex]
where [tex]v_1,v_2[/tex] are initial/final velocities, [tex]a[/tex] is acceleration, and [tex]\Delta x[/tex] is displacement. We're given [tex]a=-5.00\frac{\rm m}{\rm s^2}[/tex] and [tex]\Delta x=15.0\,\rm m[/tex], and the car comes to a stop so [tex]v_2=0[/tex]. Solve for [tex]v_1[/tex].
[tex]-{v_1}^2 = 2\left(-5.00\dfrac{\rm m}{\rm s^2}\right) (15.0\,\mathrm m) \\\\ ~~~~ \implies {v_1}^2 = 150.\dfrac{\rm m^2}{\rm s^2} \\\\ ~~~~ \implies v_1 = \sqrt{150.\dfrac{\rm m^2}{\rm s^2}} \approx \boxed{12.2 \dfrac{\rm m}{\rm s}}[/tex]
