Question

The coefficient of kinetic friction between the road and the tyres is $0.2$. Then the shortest distance in which a car be stopped if it was travelling with a velocity of $72$ kmph is ( g = 9.8 $m{s}^{-2}$)

• A. $100m$
• B. $110m$
• C. $105m$
• D. $102m$

Answer 1

Answer: (D)

$102m$

Given,

$\mu =0.2$

$g=9.8m/s^2$

$u=72km/h=\frac{72\times 1000}{60\times 60}=20m/s$

$v=0m/s$

The frictional force,

$f_{\mu }=-\mu mg=ma$

$a=-\mu g=-0.2\times 9.8=-1.96m/s^2$

From the 3rd equation of motion,

$2as=v^2-u^2$

$-2\times 1.96\times s=0^2-20\times 20$

$s=\frac{20\times 20}{2\times 1.96}=102m$

The correct option is D.