c
(c) Let car starts from point \(A\) from rest and moves up to point \(B\) with acceleration \(f\)

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

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

\(⇒\)  \(S = \frac{1}{{72}}f{t^2}\).