Q. If the velocity of a car is increased by 20%, then the minimum distance in which it can be stopped increase by
A) 44%
B) 55%
C) 66%
D) 88%
Explanation
(a) $0^{2} - v^{2} = - 2aS$ or $v^{2} \propto S;\left( v + \frac{20v}{100} \right)^{2} \propto S'$
or $\frac{S'}{S} = \left( 1 + \frac{20}{100} \right)^{2} = \frac{36}{25}$
$\therefore\ \left( \frac{S'}{S} - 1\ \right) \times 100 = \left( \frac{36}{25} - 1 \right) \times 100 = 44\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.