• # question_answer A car moving with a speed of $50\text{ }km/h$can be stopped by brakes, over a distance of $6m$. If the same car is moving at a speed of $100\text{ }km/h,$ the stopping distance is A)   $12\text{ }m$ B)  $18\text{ }m$ C)  $6\text{ }m$ D)  $24\text{ }m$

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