Question

Help! A particle moves along a straight line so that its distance to the right of the origin at time t is given by:

s(t) = 2t^3 - 6t^2 + 8

a) At what time(s) is the particle at the origin?
b) At what time(s) is the particle not moving?
c) At what time(s) is the velocity of the particle neither increasing nor decreasing; that is, at what time(s) is the particle neither accelerating nor decelerating.

Answer 1

a) set the equation equal to zero and solve and since this is
\[s(t)=2 (t-2)^2 (t+1)\] you get
\[t=2\] or
\[t=-1\] but you cannot go back in time so leave it at
\[t=2\]

Answer 2

b) i assume by "not moving" it means "changing direction' so take the derivative, set it equal to zero and solve. you get
\[s'(t)= 6t^2-12t = 6 t(t-2) \]
so
\[t=0\] or
\[t=2\]

Answer 3

and finally for c) take the second derivative, set it equal to zero and solve

Answer 4

Thank you!