Answer
$$\eqalign{
& {\bf{v}}\left( 2 \right) = 3{\bf{i}} + {\bf{j}} + {\bf{k}} \cr
& {\bf{a}}\left( 2 \right) = \frac{1}{2}{\bf{k}} \cr} $$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = 3t{\bf{i}} + t{\bf{j}} + \frac{1}{4}{t^2}{\bf{k}},{\text{ }}t = 2 \cr
& \left( {\bf{a}} \right){\text{Find the vectors: }}{\bf{v}}\left( t \right),{\text{ }}{\bf{a}}\left( t \right){\text{ and speed}}{\text{.}} \cr
& {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr
& {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left[ {3t{\bf{i}} + t{\bf{j}} + \frac{1}{4}{t^2}{\bf{k}}} \right] \cr
& {\bf{v}}\left( t \right) = 3{\bf{i}} + {\bf{j}} + \frac{1}{2}t{\bf{k}} \cr
& {\text{speed}} = \left\| {{\bf{v}}\left( t \right)} \right\| = \left\| {3{\bf{i}} + {\bf{j}} + \frac{1}{2}t{\bf{k}}} \right\| \cr
& {\text{speed}} = \sqrt {9 + 1 + \frac{1}{4}{t^2}} = \sqrt {10 + \frac{1}{4}{t^2}} \cr
& {\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right) \cr
& {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ {3{\bf{i}} + {\bf{j}} + \frac{1}{2}t{\bf{k}}} \right] \cr
& {\bf{a}}\left( t \right) = \frac{1}{2}{\bf{k}} \cr
& \cr
& \left( {\bf{b}} \right){\text{Evaluating }}{\bf{v}}\left( t \right),{\text{ }}{\bf{a}}\left( t \right){\text{ at }}t = 2 \cr
& {\bf{v}}\left( 2 \right) = 3{\bf{i}} + {\bf{j}} + \frac{1}{2}\left( 2 \right){\bf{k}} \cr
& {\bf{v}}\left( 2 \right) = 3{\bf{i}} + {\bf{j}} + {\bf{k}} \cr
& {\bf{a}}\left( 2 \right) = \frac{1}{2}{\bf{k}} \cr} $$
