In an accident site, tyre marks shows that a car was travelling along straight level street skidded for a total distance of 23 m, after the brakes applied. The coefficient of friction between the tyre and road estimated to be 0.5, what was the probable speed of car when break applied. 

Correct option is A

​$$\begin{aligned}&{ \textbf{Given:}} \\&\text{Skid distance, } s = 23\, \text{m} \\&\text{Coefficient of friction, } \mu = 0.5 \\&\text{Final speed after skidding, } v = 0 \, (\text{comes to rest}) \\&\text{Acceleration due to gravity, } g = 9.81\, \text{m/s}^2 \\[1.5em]&{ \textbf{Deceleration due to friction}} \\&a = \mu g = 0.5 \times 9.81 = 4.905\, \text{m/s}^2 \\[1.5em]&{ \textbf{Step 2: Use kinematic equation}} \\&v^2 = u^2 + 2as \Rightarrow 0 = u^2 - 2as \Rightarrow u^2 = 2as \\&u = \sqrt{2 \times 4.905 \times 23} = \sqrt{225.63} \approx \boxed{15.02\, \text{m/s}} \\[1em]&\text{Convert to km/h:} \\&15.02 \times 3.6 \approx \boxed{54.1\, \text{km/h}}\end{aligned}$$​​