• # question_answer Speeds of two identical cars are u and 4 u at a specific instant. The ratio of the respective distances at which the two cars are stopped from that instant is                                 [AIEEE 2002] A) $1:1$ B) $1:4$ C) $1:8$ D) $1:16$

 [d] In this question, the cars are identical means coefficient of friction between the tyre and the ground is same for both the cars, as a result retardation is same for both the cars equal to$\mu g$.          $(\because mg=\mu mg\Rightarrow a=\mu g)$ Let first car travel distance ${{s}_{1}}$, before stopping while second car travel distance ${{s}_{2}}$, then from ${{v}^{2}}={{u}^{2}}-2$ as, we get $O={{u}^{2}}-2\mu g\times {{s}_{1}}$ $\Rightarrow$   ${{s}_{1}}=\frac{{{u}^{2}}}{2\mu g}$ and      $0={{(4u)}^{2}}-2\mu g\times {{s}_{2}}$ $\Rightarrow$   ${{s}_{2}}=\frac{16{{u}^{2}}}{2\mu g}=16{{s}_{1}}$ $\Rightarrow$   $\frac{{{s}_{1}}}{{{s}_{2}}}=\frac{1}{16}$