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

\(\mathrm{v}=6 \mathrm{t}-\frac{\mathrm{t}^2}{6}\) 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 \(\mathrm{t}=0 \Rightarrow \mathrm{c}=0\)
\(
\therefore \mathrm{s}=3 \mathrm{t}^2-\frac{\mathrm{t}^3}{18} \Rightarrow(\mathrm{s})_{\mathrm{t}=3}=3(3)^2-\frac{(3)^3}{18}=\frac{51}{2} \text { units }
\)