The position function of a particle in rectilinear motion is given by s(t) s(t) = t3 - 9t2 + 24t + 1 for t ≥ 0 with t measured in seconds and s(t) measured in feet. Find the position and acceleration of the particle at the instant when the particle reverses direction. Include units in your answer.

Answer:[tex]s = 21\ ft\\\\a = -6\ \frac{ft}{s^2}[/tex]Step-by-step explanation:To find the changes in the slope of the function s(t) we find its derivative[tex]\frac{ds(t)}{dt} = 3t^2 - 18t +24\\[/tex]Now we make [tex]\frac{ds(t)}{dt} = 0[/tex][tex]3t^2 - 18t +24 = 0[/tex][tex]3t^2 - 18t +24\\\\3(t^2-6t + 8)\\\\3(t-4)(t-2)\\\\t_1 =4\\t_2=2\\[/tex]The particle changes direction for the first time at t = 2 secThe position at t = 2 sec is:[tex]s(t=2) = (2)^3 - 9(2)^2 + 24(2) + 1\\\\\s(2) = 21\ ft[/tex]The acceleration after t = 2 sec is the second derivative of s(t), evaluated at t = 2:[tex]\frac{d^2s(t)}{d^2t} = 6t -18\\\\\frac{d^2s(t)}{d^2t}(2) = 6(2) -18\\\\ a = 12-18\\\\a = -6\ \frac{ft}{s^2}[/tex]