Explanation

(c) Let car starts from point $A$ from rest moves up to point $B$ with acceleration $f$
Velocity of car at point $B,\ \ \ \ \ \ v = \sqrt{2fS}$
$\lbrack As\ v^{2} = u^{2} + 2as\rbrack$
Car moves distance $BC$ with this constant velocity in time $t$
$x = \sqrt{2fS}.t$ [As $s = ut$] … (i)
So the velocity of car at point $C$ also will be $\sqrt{2fS}$ and finally car stops after covering distance $y$
Distance $CD \Rightarrow y = \frac{\left( \sqrt{2fS} \right)}{2(f/2)}^{2} = \frac{2fS}{f} = 2S$ …(ii)
So, the total distance $AD = AB + BC + CD = 15S$ [Given]
$\Rightarrow S + x + 2S = 15\ S \Rightarrow x = 12S$
Substituting the value of $x$ in equation (i) we get
$x = \sqrt{2fS}.t\ \Rightarrow 12S = \ \sqrt{2fS}.t \Rightarrow 144S^{2} = 2fS.t^{2}$
$\Rightarrow S = \frac{1}{72}\ ft^{2}$