If the acceleration applied is constant, then it is the same as the average acceleration throughout the duration of the stop.
[tex]a=\dfrac{\Delta v}{\Delta t}\implies-6.4\,\dfrac{\mathrm m}{\mathrm s^2}=\dfrac{0-28\,\frac{\mathrm m}{\mathrm s}}{\Delta t}\implies\Delta t=4.4\,\mathrm s[/tex]
