Question

The position function of a particle in rectilinear motion is given by s(t) s(t) = t3 - 9t2 + 24t + 1 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

The position is 21 feet

The acceleration is -6 ft/sec²

Explanation:

∵ s(t) = t³ - 9t² + 24t + 1

- where s is the displacement in feet and t is the time in second

∵ v(t) = ds/dt ⇒ velocity is the differentiation of s(t)

∵ a(t) = d²s/dt² = dv/dt ⇒ acceleration is the differentiation of v(t)

∴ v(t) = 3t² - 18t + 24

∵ When the particle reverses means v(t) = 0

∴ 3t² - 18t + 24 = 0 ⇒ ÷ 3

∴ t² - 6t + 8 = 0

∴ (t - 2)(t - 4) = 0

∴ t - 2 = 0 ⇒ t = 2 seconds

∴ t - 4 = 0 ⇒ t = 4 seconds

- That means the particle will change its direction first after 2 seconds

and again after 4 seconds

∵ a(t) = dv/dt

∴ a(t) = 6t - 18

∵ t = 2

∴ a(2) = 6(2) - 18 = 12 - 18 = -6 ft/sec²

∵ s(t) = t³ - 9t² + 24t + 1

∴ s(2) = 2³ - 9(2)² + 24(2) + 1 = 21 feet

Answer 2

`s = 21\ ft\\\\a = -6\ (ft)/(s^2)`

Explanation:

To find the changes in the slope of the function s(t) we find its derivative

`(ds(t))/(dt) = 3t^2 - 18t +24\\`

Now we make
`(ds(t))/(dt) = 0`

`3t^2 - 18t +24 = 0`

`3t^2 - 18t +24\\\\3(t^2-6t + 8)\\\\3(t-4)(t-2)\\\\t_1 =4\\t_2=2\\`

The particle changes direction for the first time at t = 2 sec

The position at t = 2 sec is:

`s(t=2) = (2)^3 - 9(2)^2 + 24(2) + 1\\\\\s(2) = 21\ ft`

The acceleration after t = 2 sec is the second derivative of s(t), evaluated at t = 2:

`(d^2s(t))/(d^2t) = 6t -18\\\\(d^2s(t))/(d^2t)(2) = 6(2) -18\\\\ a = 12-18\\\\a = -6\ (ft)/(s^2)`