Recall that
[tex]v_f=v_0+at[/tex]
It takes the car about 3.2 s to reduce its speed from 14.6 to 10.8 m/s, since
[tex]10.8\,\dfrac{\mathrm m}{\mathrm s}=14.6\,\dfrac{\mathrm m}{\mathrm s}+\left(-1.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\implies t=3.2\,\mathrm s[/tex]
Next, recall that
[tex]\Delta x=x_f-x_0=\dfrac{v_f+v_0}2t[/tex]
Then the car undergoes a displacement of about 41 m, since
[tex]\Delta x=\dfrac{10.8\,\frac{\mathrm m}{\mathrm s}+14.6\,\frac{\mathrm m}{\mathrm s}}2(3.2\,\mathrm s)\implies\Delta x=41\,\mathrm m[/tex]
