Answer
$${\bf{v}}\left( t \right) = \left\langle { - 2t,6{t^2}} \right\rangle ,{\text{ speed}}:{\text{ }}2t\sqrt {1 + 9{t^2}} {\text{ and }}{\bf{a}}\left( t \right) = \left\langle { - 2,12t} \right\rangle $$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {1 - {t^2},3 + 2{t^3}} \right\rangle ,\,\,\,\,\,{\text{for }}t \geqslant 0 \cr
& \left( a \right){\text{find the velocity and speed }} \cr
& {\text{the velocity vector is given by }}\,\,\left( {{\text{see page 817}}} \right) \cr
& {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr
& {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle {1 - {t^2},3 + 2{t^3}} \right\rangle } \right] \cr
& {\text{differentiating each component}} \cr
& {\bf{v}}\left( t \right) = \left\langle { - 2t,6{t^2}} \right\rangle \cr
& {\text{The speed of the object is the scalar function }}\left| {{\bf{v}}\left( t \right)} \right|{\text{ then}} \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \left| {\left\langle { - 2t,6{t^2}} \right\rangle } \right| \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {{{\left( { - 2t} \right)}^2} + {{\left( {6{t^2}} \right)}^2}} \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {4{t^2} + 36{t^4}} \cr
& {\text{factoring}} \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {4{t^2}\left( {1 + 9{t^2}} \right)} \cr
& \left| {{\bf{v}}\left( t \right)} \right| = 2t\sqrt {1 + 9{t^2}} \cr
& \cr
& \left( b \right){\text{find the acceleration of the object}} \cr
& {\text{the acceleration vector is given by }}\,\,{\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right)\left( {{\text{see page 817}}} \right) \cr
& {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle { - 2t,6{t^2}} \right\rangle } \right] \cr
& {\text{differentiating each component}} \cr
& {\bf{a}}\left( t \right) = \left\langle { - 2,12t} \right\rangle \cr} $$
