The position of a particle moving in a straight line is given by the function S (t) = t3 − 6t2,t ≥ 0. The graph of S against t is shown. The corresponding velocity–time graph is alsoshown. The function describing the velocity is V(t) = 3t2 − 12t.1 2 3 41 2 3 45 650–5–10–10–20010–30S(t) = t3 – 6t2 V(t) = 3t2 – 12tVSt ta Find the average velocity of the particle for the intervals:i [3.5, 4.5] ii [3.9, 4.1] iii [3.99, 4.01]b From part a, what is the instantaneous velocity when t = 4?c i For what values of t is the velocity positive?ii For what values of t is the velocity negative

a) The average velocity of a particle over an interval [a, b] is given by the formula (S(b) - S(a)) / (b - a).

i) For the interval [3.5, 4.5], the average velocity is (S(4.5) - S(3.5)) / (4.5 - 3.5). Substituting the given function S(t) = t^3 - 6t^2, we get (4.5^3 - 64.5^2 - (3.5^3 - 63.5^2)) / 1.

ii) For the interval [3.9, 4.1], the average velocity is (S(4.1) - S(3.9)) / (4.1 - 3.9). Substituting the given function S(t), we get (4.1^3 - 64.1^2 - (3.9^3 - 63.9^2)) / 0.2.

iii) For the interval [3.99, 4.01], the average velocity is (S(4.01) - S(3.99)) / (4.01 - 3.99). Substituting the given function S(t), we get (4.01^3 - 64.01^2 - (3.99^3 - 63.99^2)) / 0.02.

b) The instantaneous velocity at t = 4 is given by the derivative of the position function at that point, which is the velocity function V(t). So, V(4) = 34^2 - 124.

c) The velocity is positive when the velocity function V(t) is greater than 0, and negative when V(t) is less than 0.

i) To find when the velocity is positive, solve the inequality 3t^2 - 12t > 0.

ii) To find when the velocity is negative, solve the inequality 3t^2 - 12t < 0.