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

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24 m

6 m

12 m

18 m

answer is A.

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Detailed Solution

$\begin{array}{l} a = \frac{u_{1}^{2}}{2 s_{1}}, a = \frac{u_{2}^{2}}{2 s_{2}} \\ \Rightarrow \frac{s_{1}}{s_{2}} = \frac{u_{2}^{2}}{u_{1}^{2}} \\ \Rightarrow s_{2} = 6 \times {(\frac{100}{50})}^{2} = 24 m \end{array}$

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