1. Acceleration phase

The car starts from rest and accelerates at rate $f$ through distance $S$.

Using $v^2=u^2+2as$ with $u=0$, $a=f$, $s=S$, $v^2=2fS$ So the speed after the first phase is $v=\sqrt{2fS}$

2. Constant speed phase

The car then moves with constant speed $v$ for time $t$.

Distance covered in this phase: $s_2=vt=t\sqrt{2fS}$

3. Deceleration phase

Now the car decelerates at rate $\dfrac{f}{2}$ and comes to rest.

Using $v_f^2=u^2+2as$ Here $v_f=0$, initial speed $u=v$, acceleration $a=-\dfrac{f}{2}$, and distance $s=s_3$: $0=v^2+2\left(-\frac{f}{2}\right)s_3$ $0=v^2-fs_3$ $s_3=\frac{v^2}{f}$ Since $v^2=2fS$, $s_3=\frac{2fS}{f}=2S$

4. Total distance condition

Total distance traversed is $15S$: $S+s_2+s_3=15S$ Substitute $s_2=t\sqrt{2fS}$ and $s_3=2S$: $S+t\sqrt{2fS}+2S=15S$ $t\sqrt{2fS}=12S$

5. Solve for $S$

Square both sides: $2fS\,t^2=144S^2$ Assuming $S\neq 0$, divide by $S$: $2ft^2=144S$ $S=\frac{2ft^2}{144}=\frac{1}{72}ft^2$

6. Match with options

$S=\frac{1}{72}ft^2$ This matches Option D.