Given acceleration a = (9 − 3t) ms⁻².

Given acceleration $a = (9 - 3t) \text{ ms}^{-2}$.
Velocity $V(t)$ is the integral of acceleration with respect to time:

$$V(t) = \int a(t)\,dt = \int (9 - 3t)\,dt$$ $$V(t) = 9t - \frac{3t^2}{2} + C$$

The initial velocity was $1 \text{ ms}^{-1}$, meaning $V(0) = 1$.

$$1 = 9(0) - \frac{3(0)^2}{2} + C$$ $$1 = C$$

$$V(t) = 9t - \frac{3t^2}{2} + 1$$

The velocity of the particle at any given time $t$ is $\boxed{\mathbf{V(t) = \left(9t - \frac{3t^2}{2} + 1\right) \text{ ms}^{-1}}}$.

$$a(t) = 9 - 3t = 0$$ $$3t = 9$$ $$t = 3 \text{ s}$$

$$V_{max} = V(3) = 9(3) - \frac{3(3)^2}{2} + 1$$ $$V_{max} = 27 - \frac{3(9)}{2} + 1$$ $$V_{max} = 27 - \frac{27}{2} + 1$$ $$V_{max} = 28 - 13.5$$ $$V_{max} = 14.5 \text{ ms}^{-1}$$

The maximum velocity of the particle is $\boxed{\mathbf{14.5 \text{ ms}^{-1}}}$.

Displacement $s(t)$ is the integral of velocity with respect to time:

$$s(t) = \int V(t)\,dt = \int \left(9t - \frac{3t^2}{2} + 1\right)\,dt$$ $$s(t) = \frac{9t^2}{2} - \frac{3t^3}{2 \times 3} + t + D$$ $$s(t) = \frac{9t^2}{2} - \frac{t^3}{2} + t + D$$

Assume the particle starts at $s=0$ when $t=0$.

$$0 = \frac{9(0)^2}{2} - \frac{(0)^3}{2} + 0 + D$$ $$0 = D$$

$$s(3) = \frac{9(3)^2}{2} - \frac{(3)^3}{2} + 3$$ $$s(3) = \frac{9(9)}{2} - \frac{27}{2} + 3$$ $$s(3) = \frac{81}{2} - \frac{27}{2} + 3$$ $$s(3) = \frac{81 - 27}{2} + 3$$ $$s(3) = \frac{54}{2} + 3$$ $$s(3) = 27 + 3$$ $$s(3) = 30 \text{ m}$$

The distance covered by the particle by the time it attained maximum velocity is $\boxed{\mathbf{30 \text{ m}}}$.