Velocity at a given time

Differentiate $s(t) = t^3 - 6t^2 + 9t$ term by term using the power rule: $v(t) = s'(t) = 3t^2 - 12t + 9$.

Substitute $t = 2$: $v(2) = 3(2)^2 - 12(2) + 9 = 12 - 24 + 9 = -3$. So the velocity at $t=2$ is $-3$ m/s.

Differentiate $v(t) = 3t^2 - 12t + 9$: $a(t) = v'(t) = 6t - 12$.

Substitute $t = 2$: $a(2) = 6(2) - 12 = 12 - 12 = 0$. So the acceleration at $t=2$ is $0$ m/s².

Since $v(2) = -3 < 0$, the particle is moving in the negative direction at $t = 2$.

Since $a(2) = 0$, the particle's speed is momentarily neither increasing nor decreasing at $t=2$ — it is at a critical point of the velocity function, so it is neither speeding up nor slowing down at that instant.