MATH SOLVE

3 months ago

Q:
# 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.

Accepted Solution

A:

Answer:[tex]s = 21\ ft\\\\a = -6\ \frac{ft}{s^2}[/tex]Step-by-step explanation:To find the changes in the slope of the function s(t) we find its derivative[tex]\frac{ds(t)}{dt} = 3t^2 - 18t +24\\[/tex]Now we make [tex]\frac{ds(t)}{dt} = 0[/tex][tex]3t^2 - 18t +24 = 0[/tex][tex]3t^2 - 18t +24\\\\3(t^2-6t + 8)\\\\3(t-4)(t-2)\\\\t_1 =4\\t_2=2\\[/tex]The particle changes direction for the first time at t = 2 secThe position at t = 2 sec is:[tex]s(t=2) = (2)^3 - 9(2)^2 + 24(2) + 1\\\\\s(2) = 21\ ft[/tex]The acceleration after t = 2 sec is the second derivative of s(t), evaluated at t = 2:[tex]\frac{d^2s(t)}{d^2t} = 6t -18\\\\\frac{d^2s(t)}{d^2t}(2) = 6(2) -18\\\\ a = 12-18\\\\a = -6\ \frac{ft}{s^2}[/tex]