AP EAPCET 24-Aug-2021 Shift 1 Paper

Given that, initial speed of car, $v_i=v$ Final speed of car when it stops, $v_f=0$ Distance travelled $=s$ Retarding force $=F$ By Newton's law of motion, $F = m a = m \left( \frac{v_i^2 - v_i^2}{2 s} \right)$ ...(i) $\begin{bmatrix} \text{using relation} \\ a = \frac{v_f^2 - v_i^2}{2 s} \end{bmatrix}$ $= \frac{m (-v^2)}{2 s}$ $F = \frac{-m v^2}{2 s}$ When, retarding force, $F' = 3 F$ Let $s'=$ New distance travelled. Similarly, from Eq. (i) we can write $3 F = \frac{-m v^2}{2 s'}$ ...(ii) Substituting value of F from Eq. (i) into Eq. (ii), we get $3 \left( \frac{-m v^2}{2 s} \right) = \frac{-m v^2}{2 s^2}$ ⇒ $s' = \frac{s}{3}$ Hence, the body will stop by travelling distance $\frac{s}{3}$, when retarding force is $3 F$.