A car starting from rest accelerates at the rate $f$ through a distance $S$, then continues at constant speed for time $t$ and then decelerates at the rate $f/2$ to come to rest. If the total distance traversed is $15S$, then:

Acceleration phase: $v^2=2fS⟹v=\sqrt{2fS}$. Time $t_1=v/f=\sqrt{2S/f}$.Constant speed phase: Distance $S_2=v\cdot t=\sqrt{2fS}\cdot t$.Deceleration phase ($a=-f/2$): $0^2=v^2+2(-f/2)S_3⟹S_3=v^2/f=(2fS)/f=2S$.Total distance $=S+S_2+S_3=15S$.
$$S+\sqrt{2fS}t+2S=15S$$
$$\sqrt{2fS}t=12S$$
Squaring both sides:
$$2fS{t}^2=144S^2$$
$$2f{t}^2=144S⟹S=\frac{2f{t}^2}{144}=\frac{f{t}^2}{72}$$
None of the given options match the calculation based on the problem statement $15S$. However, if we evaluate the intent of the options, there might be a typo in the source numericals. Following the steps, $S=\frac{1}{72}f{t}^2$.