Here the position vector of the point is given by,

 $\vec{r}=(\frac{4}{3}t^3-2t)\hat{i}+t^2\hat{j}$

So the velocity is  $\displaystyle{\vec{v}=\frac{d\vec{r}}{dt}}$

$or,\ \displaystyle{\vec{v}=\frac{d}{dt}[(\frac{4}{3}t^3-2t)\hat{i}+t^2\hat{j}]}$

$or,\ \vec{v}=\displaystyle{(\frac{4}{3}\times{3}t^2-2)\hat{i}+2t\hat{j}}$

$or,\ \vec{v}=(4t^2-2)\hat{i}+2t\hat{j}$

Now, at t=3sec.

$\vec{v}=(4\times{3}^2-2)\hat{i}+2\times{3}\hat{j}$

$or,\ \vec{v}=34\hat{i}+6\hat{j}$

So the magnitude of the velocity $\vec{v}$ is,

$v=|\vec{v}|=\sqrt{34^2+6^2}=34.5\ m/sec$

Again we know that acceleration is the time derivative of the velocity vector. So the acceleration is given by,

$\displaystyle{\vec{a}=\frac{d\vec{v}}{dt}}$

$or,\ \vec{a}=\displaystyle{\frac{d}{dt}[(4t^2-2)\hat{i}+2t\hat{j}]}$

$or,\ \vec{a}=(4\times{2}t-0)\hat{i}+2\hat{j}$

$or,\ \vec{a}=8t\hat{i}+2\hat{j}$

So at t=3 sec. the acceleration is

$\vec{a}=8\times{3}\hat{i}+2\hat{j}$

$or,\ \vec{a}=24\hat{i}+2\hat{j}$

So the magnitude of the acceleration is given by,

$a=|\vec{a}|=\sqrt{24^2+2^2}=24.1\ m/{sec}^2$