Let s(t) = t^3 - 9t^2 + 24t be the position function of a particle moving along a coordinate line where s is in meters and t is in minutes. At what time is the particle stopped and when is the particle speeding up or slowing down?

Expert Answers

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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.

Approved by eNotes Editorial Team

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