From the energy conservation principle, the loss in the kinetic energy of the car is equal to the work done by the friction force on the car.
Now, the change in kinetic energy is \(\left( {\frac{1}{2}m{v^2} - \frac{1}{2}mv_{\rm{o}}^2} \right)\).
The work done by the friction force is \(Fd\cos {180^ \circ }\) as the applied force is opposing the motion.
Now, from the work-energy theorem, you can write:
\(\begin{array}{c}\frac{1}{2}m{v^2} - \frac{1}{2}mv_{\rm{o}}^2 = Fd\cos {180^ \circ }\\\frac{1}{2}m{v^2} - \frac{1}{2}mv_{\rm{o}}^2 = {\mu _{\rm{k}}}mgd\cos {180^ \circ }\\\left( {\frac{1}{2}m \times {0^2}} \right) - \frac{1}{2}mv_{\rm{o}}^2 = 0.30 \times m \times \left( {9.80\;{\rm{m/}}{{\rm{s}}^2}} \right) \times \left( {78\;{\rm{m}}} \right)\cos {180^ \circ }\\{v_{\rm{o}}} = 21.42\;{\rm{m/s}}\\{v_{\rm{o}}} \approx 21\;{\rm{m/s}}\end{array}\)
Hence, the speed of the car is 21 m/s whenthe driver slams on the brakes.
The car's mass does not matter because both the friction force and the car's kinetic energy are proportional to the car's mass. So, the mass gets canceled during the calculation of the speed, and speed does not appear to be a function of mass.