Answer
$$\mathbf{v}(t)=\langle{2t+1, 2t-1, 3t^2}\rangle\\
\mathbf{a}(t)=\langle{2, 2, 6t}\rangle\\
|\mathbf{v}(t)|=\sqrt{(2t+1)^2+(2t-1)^2+(3t^2)^2}$$
Work Step by Step
1) Differentiate $\mathbf{r}(t)$ to find $\mathbf{r}'(t)=\mathbf{v}(t)$ using power rule.
$$\mathbf{v}(t)=\mathbf{r}'(t)=\frac{d(\langle{t^2+t, t^2-t, t^3}\rangle)}{dt}\\
\mathbf{v}(t)=\langle{2t+1, 2t-1, 3t^2}\rangle$$
2) Differentiate $\mathbf{v}(t)$ to find $\mathbf{v}'(t)=\mathbf{a}(t)$ using power rule.
$$\mathbf{a}(t)=\mathbf{v}'(t)=\frac{d(\langle{2t+1, 2t-1, 3t^2}\rangle)}{dt}\\
\mathbf{a}(t)=\langle{2, 2, 6t}\rangle$$
3) Find magnitude $|\mathbf{v}(t)|$ of $\mathbf{v}(t)$ to find speed.
$$speed=|\mathbf{v}(t)|=\sqrt{(2t+1)^2+(2t-1)^2+(3t^2)}$$
