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Amotorcarmovingwithvelocity $7 \mathrm{~m} / \mathrm{sstopsin} 10 \mathrm{mdistance}$ when brakes are applied. What is the relation between the resistance force $(\mathrm{R})$ and the weight $(\mathrm{W})$ of the car? $\left(\right.$ Take value of $g=9.8 \mathrm{~m} / \mathrm{s}^2$ )
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Verified Answer
The correct answer is:
$\mathrm{R}=-\frac{\mathrm{W}}{4}$
From equation of motion
$$
\begin{aligned}
& v^2=u^2+2 a s \\
& \Rightarrow \quad a=\frac{v^2-u^2}{2 s} \\
& =-\frac{(7)^2+(0)^2}{2 \times 10}=-\frac{49}{20}=-2.45 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$
$$
=-\frac{\mathrm{g}}{4} \mathrm{~m} / \mathrm{s}^2
$$
We have $g=9.8 \mathrm{~m} / \mathrm{s}^2$
$$
\begin{aligned}
& \mathrm{a}=-\frac{\mathrm{g}}{4} \\
& \mathrm{ma}=-\frac{\mathrm{mg}}{4} \\
& \mathrm{R}=-\frac{\mathrm{W}}{4}
\end{aligned}
$$
$$
\begin{aligned}
& v^2=u^2+2 a s \\
& \Rightarrow \quad a=\frac{v^2-u^2}{2 s} \\
& =-\frac{(7)^2+(0)^2}{2 \times 10}=-\frac{49}{20}=-2.45 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$
$$
=-\frac{\mathrm{g}}{4} \mathrm{~m} / \mathrm{s}^2
$$
We have $g=9.8 \mathrm{~m} / \mathrm{s}^2$
$$
\begin{aligned}
& \mathrm{a}=-\frac{\mathrm{g}}{4} \\
& \mathrm{ma}=-\frac{\mathrm{mg}}{4} \\
& \mathrm{R}=-\frac{\mathrm{W}}{4}
\end{aligned}
$$
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