The position function of a particle in rectilinear motion is given by s(t) = 2t^3 – 21t^2 + 60t + 3 for t ≥ 0. Find the position and acceleration of the particle at the instant the when the particle reverses direction. Include units in your answer.

Answer 1

find ds/dt
when it reverses ds/st = 0
find the acceleration d^2s/dt^2 then

Answer 2

When the particle reverses direction, this means that there is a sign change in the velocity of the particle or s'(t).

s'(t) = 6t^2 - 42t + 60

Setting it equal to 0 to find critical points;
6t^2 - 42t + 60 = 0
t^2 - 7t + 10 = 0
(t-5)(t-2) = 0
t-5 = 0
t = 5
t-2 = 0
t = 2

The two instances in which the particle reverses direction are t = 2 and t = 5.
To find the acceleration of the particle at these instances, the second derivative of the position function or the first derivative of the velocity function must be taken.

s''(t) = 12t - 42
s''(2) = 12(2) - 42
= 24 - 42
= -18 ft/s^2
s''(5) = 12(5) - 42
= 60 - 42
= 18 ft/s^2