Answer
The speed at $t=1$ is $v\left( 1 \right) \simeq 4.18$
Work Step by Step
We evaluate $\frac{{{\bf{r}}\left( {1 + h} \right) - {\bf{r}}\left( 1 \right)}}{h}$ for $h = - 0.2, - 0.1,0.1,0.2$ and list them in the following table:
$\begin{array}{*{20}{c}}
h&{{\bf{r}}\left( {1 + h} \right)}\\
{ - 0.2}&{{\bf{r}}\left( {0.8} \right) = \left( {1.557,2.459, - 1.97} \right)}\\
{ - 0.1}&{{\bf{r}}\left( {0.9} \right) = \left( {1.559,2.634, - 1.74} \right)}\\
0&{{\bf{r}}\left( 1 \right) = \left( {1.54,2.841, - 1.443} \right)}\\
{0.1}&{{\bf{r}}\left( {1.1} \right) = \left( {1.499,3.078, - 1.035} \right)}\\
{0.2}&{{\bf{r}}\left( {1.2} \right) = \left( {1.435,3.342, - 0.428} \right)}
\end{array}\begin{array}{*{20}{c}}
{{\bf{r}}\left( {1 + h} \right) - {\bf{r}}\left( 1 \right)}&{\frac{{{\bf{r}}\left( {1 + h} \right) - {\bf{r}}\left( 1 \right)}}{h}}\\
{\left( {0.017, - 0.382, - 0.527} \right)}&{\left( { - 0.085,1.91,2.635} \right)}\\
{\left( {0.019, - 0.207, - 0.297} \right)}&{\left( { - 0.19,2.07,2.97} \right)}\\
{\left( {0,0,0} \right)}&{\left( {0,0,0} \right)}\\
{\left( { - 0.041,0.237,0.408} \right)}&{\left( { - 0.41,2.37,4.08} \right)}\\
{\left( { - 0.105,0.501,1.015} \right)}&{\left( { - 0.525,2.505,5.075} \right)}
\end{array}$
From the table above we estimate the velocity at $t=1$ by taking the average of velocity at $h=-0.1$ and $h=0.1$. Thus,
${\bf{v}}\left( 1 \right) = \frac{{\left( { - 0.19,2.07,2.97} \right) + \left( { - 0.41,2.37,4.08} \right)}}{2} = \left( { - 0.3,2.22,3.525} \right)$
So, the speed at $t=1$ is
$v\left( 1 \right) = \sqrt {{\bf{v}}\left( 1 \right)\cdot{\bf{v}}\left( 1 \right)} = \sqrt {{{0.3}^2} + {{2.22}^2} + {{3.525}^2}} \simeq 4.18$
