Answer
The velocity vector:
${\bf{v}}\left( s \right) = \left( { - \frac{{2s}}{{{{\left( {1 + {s^2}} \right)}^2}}},\frac{{1 - {s^2}}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right)$
The speed at $s=2$ is $v\left( 2 \right) = \frac{1}{5}$
The acceleration vector:
${\bf{a}}\left( s \right) = {\bf{v}}'\left( s \right) = \left( {\frac{{6{s^2} - 2}}{{{{\left( {1 + {s^2}} \right)}^3}}},\frac{{2{s^3} - 6s}}{{{{\left( {1 + {s^2}} \right)}^3}}}} \right)$
The acceleration vector at $s=2$ is ${\bf{a}}\left( 2 \right) = \left( {\frac{{22}}{{125}},\frac{4}{{125}}} \right)$
Work Step by Step
We have ${\bf{r}}\left( s \right) = \left( {\frac{1}{{1 + {s^2}}},\frac{s}{{1 + {s^2}}}} \right)$. Write ${\bf{r}}\left( s \right) = \left( {{{\left( {1 + {s^2}} \right)}^{ - 1}},s{{\left( {1 + {s^2}} \right)}^{ - 1}}} \right)$.
The velocity vector:
${\bf{v}}\left( s \right) = {\bf{r}}'\left( t \right) = \left( { - \frac{{2s}}{{{{\left( {1 + {s^2}} \right)}^2}}},{{\left( {1 + {s^2}} \right)}^{ - 1}} - \frac{{2{s^2}}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right)$
${\bf{v}}\left( s \right) = \left( { - \frac{{2s}}{{{{\left( {1 + {s^2}} \right)}^2}}},\frac{{1 + {s^2} - 2{s^2}}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right)$
${\bf{v}}\left( s \right) = \left( { - \frac{{2s}}{{{{\left( {1 + {s^2}} \right)}^2}}},\frac{{1 - {s^2}}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right)$
The velocity vector at $s=2$, ${\bf{v}}\left( 2 \right) = \left( { - \frac{4}{{25}}, - \frac{3}{{25}}} \right)$
So, the speed at $s=2$ is
$v\left( 2 \right) = \sqrt {{\bf{v}}\left( 2 \right)\cdot{\bf{v}}\left( 2 \right)} = \sqrt {{{\left( { - \frac{4}{{25}}} \right)}^2} + {{\left( { - \frac{3}{{25}}} \right)}^2}} = \frac{1}{5}$
From the above result we obtain
${\bf{v}}\left( s \right) = \left( { - \frac{{2s}}{{{{\left( {1 + {s^2}} \right)}^2}}},\frac{{1 - {s^2}}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right)$.
Evaluate $\frac{d}{{ds}}\left( { - \frac{{2s}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right)$
$\frac{d}{{ds}}\left( { - \frac{{2s}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right) = - \frac{{2{{\left( {1 + {s^2}} \right)}^2} - \left( {2s} \right)4s\left( {1 + {s^2}} \right)}}{{{{\left( {1 + {s^2}} \right)}^4}}}$
$\frac{d}{{ds}}\left( { - \frac{{2s}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right) = - \frac{{\left( {1 + {s^2}} \right)\left( {2\left( {1 + {s^2}} \right) - 8{s^2}} \right)}}{{{{\left( {1 + {s^2}} \right)}^4}}}$
$\frac{d}{{ds}}\left( { - \frac{{2s}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right) = - \frac{{ - 6{s^2} + 2}}{{{{\left( {1 + {s^2}} \right)}^3}}} = \frac{{6{s^2} - 2}}{{{{\left( {1 + {s^2}} \right)}^3}}}$
Evaluate $\frac{d}{{ds}}\left( {\frac{{1 - {s^2}}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right)$
$\frac{d}{{ds}}\left( {\frac{{1 - {s^2}}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right) = \frac{{ - 2s{{\left( {1 + {s^2}} \right)}^2} - \left( {1 - {s^2}} \right)4s\left( {1 + {s^2}} \right)}}{{{{\left( {1 + {s^2}} \right)}^4}}}$
$\frac{d}{{ds}}\left( {\frac{{1 - {s^2}}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right) = \frac{{\left( {1 + {s^2}} \right)\left( { - 2s\left( {1 + {s^2}} \right) - 4s\left( {1 - {s^2}} \right)} \right)}}{{{{\left( {1 + {s^2}} \right)}^4}}}$
$\frac{d}{{ds}}\left( {\frac{{1 - {s^2}}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right) = \frac{{ - 2s - 2{s^3} - 4s + 4{s^3}}}{{{{\left( {1 + {s^2}} \right)}^3}}}$
$\frac{d}{{ds}}\left( {\frac{{1 - {s^2}}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right) = \frac{{2{s^3} - 6s}}{{{{\left( {1 + {s^2}} \right)}^3}}}$
Thus, the acceleration vector:
${\bf{a}}\left( s \right) = {\bf{v}}'\left( s \right) = \left( {\frac{{6{s^2} - 2}}{{{{\left( {1 + {s^2}} \right)}^3}}},\frac{{2{s^3} - 6s}}{{{{\left( {1 + {s^2}} \right)}^3}}}} \right)$
The acceleration vector at $s=2$ is ${\bf{a}}\left( 2 \right) = \left( {\frac{{22}}{{125}},\frac{4}{{125}}} \right)$
