Velocity of car at point $B$, $v = \sqrt {2fS} $

$As \,{v^2} = {u^2} + 2as]$

Car moves distance $BC$ with this constant velocity in time $t$

$x = \sqrt {2fS} \,.\,t$ ......(i) [As $s = ut$]

So the velocity of car at point $C$ also will be $\sqrt {2fs} $ and finally car stops after covering distance y.

Distance $CD ⇒y = \frac{{{{(\sqrt {2fS} )}^2}}}{{2(f/2)}}$$ = \frac{{2fS}}{f} = 2S$....(ii) $[{\rm{As }}{v^{\rm{2}}} = {u^2} - 2as\, \Rightarrow \,s = {u^2}/2a]$

So, the total distance $AD$ = $AB + BC + CD=15S$ (given)

$⇒$ $S + x + 2S = 15S$ $⇒ x = 12S$

Substituting the value of x in equation (i) we get

$x = \sqrt {2fS} \,.\,t ⇒ 12S = \sqrt {2fS} .t ⇒ 144{S^2} = 2fS.{t^2}$

$⇒$  $S = \frac{1}{{72}}f{t^2}$.