Question

Consider a car moving along a straight horizontal road with a speed of $72 \, km \, h^{- 1}$ . 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 (taking $g=10 \, m \, s^{- 2}$ )

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

Answer 1

Answer: (B)

Here $u=72kmh^{- 1}=20ms^{- 1};v=0;$
$a=-\mu g=-0.5\times 10=-5ms^{- 2}$
As $v^{2}=u^{2}+2 \, as,$
$\therefore \, \, \, s=\frac{\left(v^{2} - u^{2}\right)}{2 a}=\frac{\left(\right. 0 - \left(20\right)^{2} \left.\right)}{2 \times \left(- 5\right)}=40m$