Calculus help needed

Question

The function 𝑠(𝑡) = 6𝑡^3 − 8𝑡 + 3 describes the motion of a particle

on a line.

Part 1: Find the velocity function of the particle at any time 𝑡 ≥ 0.

Part 2: Identify the intervals when the particle is moving in the positive

direction.

Part 3: Identify the intervals when the particle is moving in the negative

direction.

Part 4: Identify the time(s) at which the particle changes direction.

Answer 1

Hi

Part 1: Find the velocity function of the particle at any time 𝑡 ≥ 0.

By definition:

velocity is ds(t)/dt=18t^2-8

Part 2: Identify the intervals when the particle is moving in the positive direction.

the particle move in the positive direction where velocity is positive : ds(t)/dt>0 => 18t^2-8>0 => t^2>8/18

=> |t|>2/3 => t>2/3 or t<-2/3 => t ∈ ]2/3,+∞[ or t ∈ ]-∞,-2/3[

Part 3: Identify the intervals when the particle is moving in the negative direction.

the same logic as the precedent question but for : ds(t)/dt<0 => |t|<2/3 => t∈ ]-2/3,2/3[

Part 4: Identify the time(s) at which the particle changes direction.

The particle change the direction when its velocity change from negative to positive or from positive to negative , and it is clear from precedent responses that it happens at t=-2/3 or at t=2/3

Good Luck