The position function of a particle in rectilinear motion is given by s(t) = 2t3 – 21t2 + 60t + 3 for t ≥ 0 with t measured in seconds and s(t) measured in feet. find the position and acceleration of the particle at the two instants when the particle reverses direction. include units in your answer.

When the direction is reversed  velocity = 0 

v = ds/dt =  6t^2 - 42t + 60 = 0

t^2 - 7t + 10 = 0

(t - 5)(t - 2) = 0

t = 2 s  and 5 s   which are the times when the direction is reversed
Position at t = 2 =  2(2)^3 - 21(2)^2 +60(2) + 3 =  55 feet
  at t = 5  position =  28 feet

Acceleration  = dv/dt = 12t - 42

At t = 2  acceleration =  -18 ft s-2
At  t = 5       = 18 ft s-2

Do you have answer choices?