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A three-wheeler starts from rest, accelerates uniformly with $1 \mathrm{~ms}^{-2}$ on a straight road for $10 \mathrm{~s}$, and then moves with uniform velocity. Plot the distance covered by the vehicle during the $\mathrm{n}^{\text {th }}$ second $(n=1,23 \ldots)$ versus $n$. What do you expect this plot to be during accelerated motion : a straight line or a parabola?
Solution:
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Verified Answer
Since $S_{n^{t h}}=u+\frac{1}{2} a(2 n-1)$ when $u=0, a=1 \mathrm{~ms}^{-2}$
$$
\begin{aligned}
&\therefore S_{n^{t h}}=0+\frac{1}{2}(2 n-1)=\frac{1}{2}(2 n-1) \\
&\therefore \text { For } n=1,2,3 \ldots \ldots \ldots \ldots \\
&S_1=\frac{1}{2}(2 \times 1-1)=0.5 \mathrm{~m}
\end{aligned}
$$
$S_2=\frac{1}{2}(2 \times 2-1)=1.5 \mathrm{~m}$
$S_3=\frac{1}{2}(2 \times 3-1)=2.5 \mathrm{~m}$
$S_4=\frac{1}{2}(2 \times 4-1)=3.5 \mathrm{~m}$
$$
\begin{aligned}
&S_5=\frac{1}{2}(2 \times 5-1)=4.5 \mathrm{~m} \\
&S_6=\frac{1}{2}(2 \times 6-1)=5.5 \mathrm{~m} \\
&S_7=\frac{1}{2}(2 \times 7-1)=6.5 \mathrm{~m} \\
&S_8=\frac{1}{2}(2 \times 8-1)=7.5 \mathrm{~m} \\
&S_9=\frac{1}{2}(2 \times 1-9)=8.5 \mathrm{~m} \\
&S_{10}=\frac{1}{2}(2 \times 10-1)=9.5 \mathrm{~m}
\end{aligned}
$$
$$
\begin{aligned}
&\therefore S_{n^{t h}}=0+\frac{1}{2}(2 n-1)=\frac{1}{2}(2 n-1) \\
&\therefore \text { For } n=1,2,3 \ldots \ldots \ldots \ldots \\
&S_1=\frac{1}{2}(2 \times 1-1)=0.5 \mathrm{~m}
\end{aligned}
$$
$S_2=\frac{1}{2}(2 \times 2-1)=1.5 \mathrm{~m}$
$S_3=\frac{1}{2}(2 \times 3-1)=2.5 \mathrm{~m}$
$S_4=\frac{1}{2}(2 \times 4-1)=3.5 \mathrm{~m}$
$$
\begin{aligned}
&S_5=\frac{1}{2}(2 \times 5-1)=4.5 \mathrm{~m} \\
&S_6=\frac{1}{2}(2 \times 6-1)=5.5 \mathrm{~m} \\
&S_7=\frac{1}{2}(2 \times 7-1)=6.5 \mathrm{~m} \\
&S_8=\frac{1}{2}(2 \times 8-1)=7.5 \mathrm{~m} \\
&S_9=\frac{1}{2}(2 \times 1-9)=8.5 \mathrm{~m} \\
&S_{10}=\frac{1}{2}(2 \times 10-1)=9.5 \mathrm{~m}
\end{aligned}
$$
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