Question

If the coefficient of static friction between the tyres and road is $0.5$, what is the shortest distance in which an automobile can be stopped when travelling at $72km/h$?

• A. 50 m
• B. 60 m
• C. 40.8 m
• D. 80.16 m

Answer 1

Answer: (C)

Frictional force acting between road and lyres retards the motion of automobile.
There is a static friction between tyres and road, so frictional force cause the retardation in velocity of a automobile.
Free body diagram of automobile is shown. From Newton's third law

$F=f_s=\mu R=\mu g$
where $m$ is the mass of automobile.
Also, $F=ma$
$ma=\mu mg$
$\Rightarrow a=$ retardation
$=\mu g$
$=0.5g$
Let automobile stops at a distance $x$, then from equation of motion
$v^2=u^2-2ax$
Given, $v=0,u=72km/h=72\times \frac{5}{18}m/s$
$=20m/s$
$g=9.8m/s^2$
$\therefore 0^2=(20{)}^2-2\times 0.5\times 9.8x$
$\Rightarrow x=\frac{20\times 20}{2\times 0.5\times 9.8}=40.8m$