Q. A car moving with a speed of $50\ {kmh}^{- 1},$ can be stopped by brakes after at least 6 m. if the same car is moving at a speed of $100\ {kmh}^{- 1},$ the minimum stopping distance is

A) 12 m
B) 18 m
C) 24 m
D) 6 m
Explanation

(c) ${\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v}^{2} = u^{2} + 2as$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 = \left( 50 \times \frac{5}{18} \right)^{2} + 2a \times 6$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a = - 16\ {ms}^{- 2}\ \ \ \ \ \ \ \ (a = retardation)$
Again, $v^{2} = u^{2} + 2as$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 = \left( 100 \times \frac{5}{18} \right)^{2} - 16 \times 2 \times s$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s = \frac{(100\ \times 5)^{2}}{18 \times 18 \times 32} = 24.1 \approx 24\ m$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.