Laws Of Motion Ques 52
52. Consider a car moving along a straight horizontal road with a speed of $72 km / h$. 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 $g=10$ $m / s^{2}$ )
[1992]
(a) $30 m$
(b) $40 m$
(c) $72 m$
(d) $20 m$
Show Answer
Answer:
Correct Answer: 52.(b)
Solution:
- (b) Here $u=72 km / h=20 m / s ; v=0$;
$a=-\mu g=-0.5 \times 10=-5 m / s^{2}$
As $v^{2}=u^{2}+2 a s$,
$\therefore s=\frac{(v^{2}-u^{2})}{2 a}=\frac{(0-(20)^{2}.}{2 \times(-5)}=40 m$
BETA
