You need to evaluate derivative of the function `s(t)` to check when the particle speeds up or slows down or even stops.
Hence, differentiating `s(t)` with respect to t yields:
`s'(t) = (t^3 - 9t^2 + 24t)'`
`s'(t) = 3t^2 - 18t + 24`
You need to check when the particle stops, hence, you need to solve the equation `s'(t) = 0` such that:
`3t^2 - 18t + 24 = 0`
Dividing by 3 yields:
`t^2 - 6t + 8 = 0`
You may use quadratic formula such that:
`t_(1,2) = (6+-sqrt(36-32))/2 => t_(1,2) = (6+-sqrt4)/2`
`t_(1,2) = (6+-2)/2 => t_1 = 4 ; t_2 = 2`
Hence, the particle stops at the t = 2 minutes and t = 4 minutes.
Since the values of s'(t) are negative over the interval `[2,4], ` hence, the particle is slowing down if `t in [2,4].`
Since the values of s'(t) are positive over the interval `[0,2]` and `[4,oo],` hence, the particle spees up if `t in [0,2]U[4,oo].`
Hence, evaluating the motion of the particle given by the function s(t), yields that the particle is speeding up for `t in [0,2]U[4,oo], ` it is slowing down for `t in [2,4]` and it stops at `t = 2` minutes and `t = 4` minutes.