A particle moves in the $\mathrm{xy}$ plane and at time $\mathrm{t}$ it is at the point whose coordinates are $({\mathrm{t}}^2, {\mathrm{t}}^3 - 2\mathrm{t}).$  Then at what instant of time will its velocity and acceleration vectors be perpendicular to each other? (Velocity, $\mathrm{v} = \frac{d\mathrm{r}}{\mathrm{dt}}$ and acceleration, $\mathrm{a} = \frac{\mathrm{dv}}{\mathrm{dt}}$)

$$\begin{gathered} \overrightarrow{\mathrm{r}}={\mathrm{t}}^2 \hat{\mathrm{i}} +\left({\mathrm{t}}^3-2\mathrm{t}\right) \hat{\mathrm{j}}, \\ \overrightarrow{\mathrm{v}}=\frac{\mathrm{d}\overrightarrow{\mathrm{r}}}{\mathrm{dt}}= 2\mathrm{t}\hat{\mathrm{i}} +\left(3{\mathrm{t}}^2-2\right) \hat{\mathrm{j}} \end{gathered}$$

$\overrightarrow{\mathrm{a}} = \frac{\mathrm{d}\overrightarrow{\mathrm{v}}}{\mathrm{dt}}= 2\hat{\mathrm{i}}+\left(6\mathrm{t}\right) \hat{\mathrm{j}}$

$$\begin{gathered} \overrightarrow{\mathrm{a}}\bullet \overrightarrow{\mathrm{v}} = \left[2\hat{\mathrm{i}}+ 6\mathrm{t}\hat{\mathrm{j}}\right]\bullet \left[2\mathrm{t}\hat{\mathrm{i}}+ \left(3{\mathrm{t}}^2-2\right) \hat{\mathrm{j}}\right] = 4\mathrm{t} +6\mathrm{t}\left(3{\mathrm{t}}^2-2\right) = 0 \\ 4\mathrm{t} + 18{\mathrm{t}}^3 - 12\mathrm{t} =0 \\ \Rightarrow 18{\mathrm{t}}^3-8\mathrm{t} = 0 \\ \Rightarrow \mathrm{t}(18{\mathrm{t}}^2-8\mathrm{t}) = 0 \\ \Rightarrow 18{\mathrm{t}}^2 = 8\mathrm{t} \mathrm{or} \mathrm{t} = 0 \\ \Rightarrow 9{\mathrm{t}}^2 = 4, \\ \Rightarrow {\mathrm{t}}^2 =\frac{4}{9}, \\ \Rightarrow \mathrm{t} =\pm \frac{2}{3} \\ \Rightarrow \mathrm{t}= \frac{2}{3} \mathrm{s} \end{gathered}$$