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

`(\mathrm ds)/(\mathrm dt)=3t^2-24t+45`

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

`(\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

`(\mathrm d)/(\mathrm dt)\left[(\mathrm ds)/(\mathrm dt)\right]=(\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.