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A car, moving with a speed of $50 \mathrm{~km} / \mathrm{hr}$, can be stopped by brakes after at least $6 \mathrm{~m}$. If the same car is moving at a speed of $100 \mathrm{~km} / \mathrm{hr}$, the minimum stopping distance is
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Verified Answer
The correct answer is:
$24 \mathrm{~m}$
$24 \mathrm{~m}$
Energy $=$ Work done by force $(\mathrm{F})$
$$
\Rightarrow \frac{1}{2} \mathrm{~m} .(50)^2=(\mathrm{F})(6) \quad \Rightarrow \mathrm{F}=\frac{2500 \mathrm{~m}}{2 \times 6}
$$
For $\mathrm{v}=100 \mathrm{~km} / \mathrm{hr} \frac{1}{2} \cdot \mathrm{m}(100)^2=(\mathrm{F})(\mathrm{S})$
$$
\begin{aligned}
& \Rightarrow \frac{1}{2} \mathrm{~m}(100)^2=\left(\frac{2500 \mathrm{~m}}{2 \times 6}\right) \mathrm{S} \\
& \Rightarrow \mathrm{S}=\frac{100 \times 100 \times 6 \times 2}{2500 \times 2}=24 \mathrm{~m}
\end{aligned}
$$
$$
\Rightarrow \frac{1}{2} \mathrm{~m} .(50)^2=(\mathrm{F})(6) \quad \Rightarrow \mathrm{F}=\frac{2500 \mathrm{~m}}{2 \times 6}
$$
For $\mathrm{v}=100 \mathrm{~km} / \mathrm{hr} \frac{1}{2} \cdot \mathrm{m}(100)^2=(\mathrm{F})(\mathrm{S})$
$$
\begin{aligned}
& \Rightarrow \frac{1}{2} \mathrm{~m}(100)^2=\left(\frac{2500 \mathrm{~m}}{2 \times 6}\right) \mathrm{S} \\
& \Rightarrow \mathrm{S}=\frac{100 \times 100 \times 6 \times 2}{2500 \times 2}=24 \mathrm{~m}
\end{aligned}
$$
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