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A car starts at time $t=0$ from an initial speed of $10 \mathrm{~m} / \mathrm{s}$ and accelerates uniformly with $2 \mathrm{~m} / \mathrm{s}^2$ on a straight road for time $0 \leq t \leq 10 \mathrm{~s}$. Let $S_1$ and $S_2$ be the distance covered by the car in time $3 \leq t \leq 4 s$ and $4 \leq t \leq 5 s$ respectively. The ratio $\frac{S_2}{S_1}$ is
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The correct answer is:
$\frac{19}{17}$
$\begin{aligned}
& |\mathrm{V}|_{\mathrm{t}=3 \mathrm{sec}}=10+2 \times 3=16 \mathrm{~m} / \mathrm{s} \\
& \mathrm{S}_{\mathrm{l}}=\left.\mathrm{S}\right|_{\mathrm{t}=3 \rightarrow 4}=16 \times 1+\frac{1}{2} \times 2 \times 1^2=17 \mathrm{~m} \\
& \text { and, }|\mathrm{V}|_{\mathrm{t}=4 \mathrm{sec}}=10+2 \times 4=18 \mathrm{~m} / \mathrm{s} \\
& \mathrm{S}_2=\left.\mathrm{S}\right|_{\mathrm{t}=4 \rightarrow 5}=18 \times 1+\frac{1}{2} \times 2 \times 1^2=19 \mathrm{~m}
\end{aligned}$
So, $\frac{\mathrm{S}_2}{\mathrm{~S}_{\mathrm{l}}}=\frac{19}{17}$
& |\mathrm{V}|_{\mathrm{t}=3 \mathrm{sec}}=10+2 \times 3=16 \mathrm{~m} / \mathrm{s} \\
& \mathrm{S}_{\mathrm{l}}=\left.\mathrm{S}\right|_{\mathrm{t}=3 \rightarrow 4}=16 \times 1+\frac{1}{2} \times 2 \times 1^2=17 \mathrm{~m} \\
& \text { and, }|\mathrm{V}|_{\mathrm{t}=4 \mathrm{sec}}=10+2 \times 4=18 \mathrm{~m} / \mathrm{s} \\
& \mathrm{S}_2=\left.\mathrm{S}\right|_{\mathrm{t}=4 \rightarrow 5}=18 \times 1+\frac{1}{2} \times 2 \times 1^2=19 \mathrm{~m}
\end{aligned}$
So, $\frac{\mathrm{S}_2}{\mathrm{~S}_{\mathrm{l}}}=\frac{19}{17}$
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