Answer
Velocity:
$$
\begin{aligned}
\mathbf{v}(t) &=\left\langle 1+\ln t, 1, -e^{-t} \right\rangle
\end{aligned}
$$
Acceleration:
$$
\begin{aligned}
\mathbf{a}(t)&=\left\langle \frac{1}{t}, 0, e^{-t} \right\rangle
\end{aligned}
$$
Speed:
$$
\begin{aligned}
|\mathbf{v}(t)| =\sqrt{2+2 \ln t+(\ln t)^{2}+e^{-2 t}}.
\end{aligned}
$$
Work Step by Step
We have the position function:
$$
\begin{aligned}
\mathbf{r}(t)&=t \ln t \mathbf{i}+t \mathbf{j}+e^{-t} \mathbf{k} \\
&=\left\langle t \ln t, t, e^{-t} \right\rangle
\end{aligned}
$$
The velocity is found by deriving the position:
$$
\begin{aligned}
\mathbf{v}(t) &=\mathbf{r}^{\prime}(t)\\
&=(1+\ln t) \mathbf{i}+\mathbf{j}-e^{-t} \mathbf{k}\\
&=\left\langle 1+\ln t, 1, -e^{-t} \right\rangle
\end{aligned}
$$
The acceleration is found by deriving the velocity:
$$
\begin{aligned}
\mathbf{a}(t)&=\mathbf{v}^{\prime}(t)\\
&=\frac{1}{t} \mathbf{i}+e^{-t} \mathbf{k}\\
&=\left\langle \frac{1}{t}, 0, e^{-t} \right\rangle
\end{aligned}
$$
The speed is the scalar version of velocity:
$$
\begin{aligned}
|\mathbf{v}(t)| &=\sqrt{(1+\ln t)^{2}+1^{2}+\left(-e^{-t}\right)^{2}} \\
&=\sqrt{2+2 \ln t+(\ln t)^{2}+e^{-2 t}}.
\end{aligned}
$$
