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 1. When the speed of a car is 𝑢, the minimum distance over which it can be stopped is 𝑠. If the speed becomes 𝑛𝑢, what will be the minimum distance over which it can be stopped during the same time? 

a) 𝑠/𝑛 b) 𝑛𝑠 c) 𝑠/𝑛 2 d) 𝑛 2 𝑠 

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1. Solution: Ans.: (d) 

WAY1✔

𝑣 2 − 𝑢 2 = 2𝑎𝑠, 𝑣 = 0 

𝑠 ∝ 𝑢 2 When the initial velocity is made 𝑛 times, the distance over which it can be stopped becomes 𝑛 2 times

WAY2✔

$$\begin{gathered} 0-v^2=2as\Rightarrow a=\frac{-v^2}{2s} \\ 0-{(nv)}^2=2ad\Rightarrow -{(nv)}^2=2\times \frac{-v^2}{2s}\times d \\ \Rightarrow d=n^{2}s \end{gathered}$$

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