Question

Consider a car moving along a straight horizontal road with a speed of $72kmph$. If the coefficient of static friction between the tyres and the road is $0.5$, the shortest distance in which the car can be stopped is $(\text{Take}g=10\frac{m}{s^2})$

• A. $30m$
• B. $40m$
• C. $72m$
• D. $20m$

Answer 1

The correct option is B $40m$

Given
$u=72kmph=20m/s$;
For stopping the car
Final velocity $v=0$;

As only friction is acting
$a=\frac{f}{m}=\frac{-\mu mg}{m}=-\mu g=-0.5\times 10=-5\frac{m}{s^2}$

Using equation of kinematics,
$v^2=u^2+2as$
$\therefore s=\frac{(v^2-u^2)}{2a}=\frac{\left(0\right)-(20{)}^2}{2\times (-5)}=40m$