## In This post you will get answer of Q.

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 :

Solution:

## 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. Let first car travel distance $s_{1}$, before stopping while second car travel distance $s_{2}$, then from

$v^{2}=u^{2}-2 a s$

$Rightarrow quad 0=u^{2}-2 mu g times s_{1}$

$Rightarrow s_{1}=frac{u^{2}}{2 mu g}$

and $0=(4 u)^{2}-2 mu g times s_{2}$

$Rightarrow s_{2}=frac{16 u^{2}}{2 mu g}=16 s_{1}$

$Rightarrow frac{s_{1}}{s_{2}}=frac{1}{16}$