Question

A car moving with a speed of $50 \,km\, h^{-1}$ can be stopped by applying brakes after moving atleast $6 \,m$. If the same car is moving at a speed of $100 \,km\, h^{-1}$ the minimum stopping distance is

• A. $12\,m$
• B. $18\,m$
• C. $6\,m$
• D. $24\,m$

Answer 1

Answer: (D)

$24\,m$

Answer 2

Case I. $v^{2}=u^{2}-2 a s$
$0=(50)^{2}-2 \times a \times \frac{6}{1000}$
$\frac{2 \times a \times 6}{1000}=50 \times 50$
$a=\frac{50 \times 50 \times 1000}{2 \times 6}$
Case II. $v^{2}=u^{2}-2 a s$
$0=(100)^{2}-2 \times \frac{50 \times 50 \times 1000}{2 \times 6} \times 5$
$s=\frac{100 \times 100 \times 6}{50 \times 50 \times 1000}$
$s=\frac{24}{1000} km$
$s=24 \,m$