Let car starts from point A from rest and moves up to point B with acceleration f

Velocity of car at point B , \(v = \sqrt{2 g S}\) \(\left[ v^{2} = u^{2} + 2as \right]\)

Car moves distance BC with this constant velocity in time t

\(x = \sqrt{2 f S t}\) ......(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 ⇒ ⇒ \(y = \frac{\left(\sqrt{2fs}\right)^2}{2\left(f/2\right)}\) \(= \frac{2 f S}{f} = 2 S\) ....(ii)

\([Asv^{2} = u^{2} - 2as \Rightarrow s = \frac{u^{2}}{2a}]\)

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

⇒ ⇒ S + X + 2S = 15S ⇒ ⇒ x = 12s

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

\(x = \sqrt{2 f S t}\) ⇒ ⇒ \(12S = \sqrt{2fSt}\) ⇒ ⇒ 144s^{2} = 27s t^{2}

⇒ ⇒ \(S = \frac{1}{72} \text{ ft}^{2}\) .