A particle moves along a horizontal path, such that its velocity is given by $v=\left(3t^2-6t\right)\mathrm{m}/\mathrm{s}$ where $t$ is the time in seconds. If it is initially located at the origin $O$, determine the distance travelled by the particle in time interval from $t=0$ to $t=3.5 \mathrm{s}$ and the particle's average velocity and average speed during the same time interval.

$\Rightarrow v=0$at $t=2 s$
For $t<2 s$, velocity is negative. At $t=2 s$, velocity is zero and for $t=2 s$ velocity is positive.

$$\begin{gathered} s_1={\int }_0^{15}vdt={\int }_0^{1s}\left(3t^2-6t\right)dt \\ =6.125 \mathrm{m} \\ =\text{ displacement upto }3.5 \mathrm{s} \\ {s}_2={\int }_0^{2}vdt={\int }_0^2\left(3t^2-6t\right)dt \\ =-4 \mathrm{m} \\ =\text{ displacement upto }2\text{ s } \end{gathered}$$

$$\begin{gathered} \therefore d=\text{ distance travelled in }3.5 \mathrm{s} \\ =4+4+6.125 \\ =14.125 \mathrm{m} \\ \text{ Average speed }=\frac{d}{t}=\frac{14.125}{3.5} \\ =4.03 \mathrm{m}/\mathrm{s} \\ \text{ Average velocity }=\frac{s_1}{t}=\frac{6.125}{3.5} \\ =1.75 \mathrm{m}/\mathrm{s}. \end{gathered}$$