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 $72\,km/h$?

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

Answer 1

Answer: (C)

40.8 m

Answer 2

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