Laws of Motion - Result Question 50

50. 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) $20 m$

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Answer:

Correct Answer: 50. (b)

Solution:

  1. (b) Here $u=72 km / h=20 m / s ; v=0$;

$a=-\mu g=-0.5 \times 10=-5 m / s^{2}$

As $v^{2}=u^{2}+2 a s$,

$\therefore s=\frac{(v^{2}-u^{2})}{2 a}=\frac{(0-(20)^{2}.}{2 \times(-5)}=40 m$