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The velocity of a particle at time $t$ is given by the relation $v=6 \mathrm{t}-\frac{\mathrm{t}^2}{6}$. Its displacement $\mathrm{S}$ is zero at $\mathrm{t}=0$, then the distance travelled in $3 \mathrm{sec}$ is
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The correct answer is:
$\frac{51}{2}$ units
$\mathrm{v}=6 \mathrm{t}-\frac{\mathrm{t}^2}{6}$ and we know that $\mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}$
$$
\begin{aligned}
& \therefore \int \mathrm{d} s=\int\left(6 \mathrm{t}-\frac{\mathrm{t}^2}{6}\right) \mathrm{dt} \\
& \therefore \mathrm{s}=\frac{6 \mathrm{t}^2}{2}-\frac{\mathrm{t}^3}{6(3)}+\mathrm{c} \Rightarrow \mathrm{s}=3 \mathrm{t}^2-\frac{\mathrm{t}^3}{18}+\mathrm{c}
\end{aligned}
$$
We know that $\mathrm{s}=0$, when $\mathrm{t}=0 \Rightarrow \mathrm{c}=0$
$$
\therefore \mathrm{s}=3 \mathrm{t}^2-\frac{\mathrm{t}^3}{18} \Rightarrow(\mathrm{s})_{\mathrm{t}=3}=3(3)^2-\frac{(3)^3}{18}=\frac{51}{2} \text { units }
$$
$$
\begin{aligned}
& \therefore \int \mathrm{d} s=\int\left(6 \mathrm{t}-\frac{\mathrm{t}^2}{6}\right) \mathrm{dt} \\
& \therefore \mathrm{s}=\frac{6 \mathrm{t}^2}{2}-\frac{\mathrm{t}^3}{6(3)}+\mathrm{c} \Rightarrow \mathrm{s}=3 \mathrm{t}^2-\frac{\mathrm{t}^3}{18}+\mathrm{c}
\end{aligned}
$$
We know that $\mathrm{s}=0$, when $\mathrm{t}=0 \Rightarrow \mathrm{c}=0$
$$
\therefore \mathrm{s}=3 \mathrm{t}^2-\frac{\mathrm{t}^3}{18} \Rightarrow(\mathrm{s})_{\mathrm{t}=3}=3(3)^2-\frac{(3)^3}{18}=\frac{51}{2} \text { units }
$$
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