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

Velocity of car at point B , \(\nu = \sqrt{2/S}\) v^2 = u^2 + 2as

Car moves distance BC with this constant velocity in time t

\(x = \sqrt{2 \beta t}\) ......(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 ⇒ ⇒ \(y \frac{\left(\sqrt{2/\beta}\right)^3}{2(f/2)}\) \(= \frac{2fs}{f} = 25\) ....(ii)

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

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

⇒ ⇒ 5 + x + 25 = 155 ⇒ ⇒ x = 125

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

\(x = \sqrt{2 \beta} \, t\) ⇒ ⇒ \(12S = \sqrt{2fS \cdot t}\) ⇒ ⇒ \(1445^{2} = 2 \beta r^{2}\)

⇒ ⇒ \(\zeta - \frac{1}{72} f t^{2}\) .