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,

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