The position function of a particle in rectilinear motion is given by s(t) = t3 – 12t2 + 45t + 4 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.

With position function[tex]s(t)=t^3-12t^2+45t+4[/tex]the velocity of the particle is given by[tex]\dfrac{\mathrm ds}{\mathrm dt}=3t^2-24t+45[/tex]The particle changes direction when its velocity changes sign. We have[tex]\dfrac{\mathrm ds}{\mathrm dt}=0\implies t=3,t=5[/tex]so the particle's direction is reversed at two different times.The acceleration is given by[tex]\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac{\mathrm ds}{\mathrm dt}\right]=\dfrac{\mathrm d^2s}{\mathrm dt^2}=6t-24[/tex]When [tex]t=3[/tex], the particle's position is [tex]s(3)=58[/tex] ft and its acceleration is [tex]s''(3)=-6[/tex] ft/s^2.When [tex]t=5[/tex], the particle's position is [tex]s(5)=54[/tex] ft and its acceleration is [tex]s''(5)=6[/tex] ft/s^2.