Question

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.

Answer 1

With position function

$s(t)=t^3-12t^2+45t+4$

the velocity of the particle is given by

$\dfrac{\mathrm ds}{\mathrm dt}=3t^2-24t+45$

The particle changes direction when its velocity changes sign. We have

$\dfrac{\mathrm ds}{\mathrm dt}=0\implies t=3,t=5$

so the particle's direction is reversed at two different times.

The acceleration is given by

$\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac{\mathrm ds}{\mathrm dt}\right]=\dfrac{\mathrm d^2s}{\mathrm dt^2}=6t-24$

When $t=3$, the particle's position is $s(3)=58$ ft and its acceleration is $s''(3)=-6$ ft/s^2.

When $t=5$, the particle's position is $s(5)=54$ ft and its acceleration is $s''(5)=6$ ft/s^2.