A particle moves along a line in such a way that its position at time tis given by s= t^3- 6t^2 + 9t+ 3 Its direction of motion changes when

Question

A particle moves along a line in such a way that its position at time tis given by s= t³- 6t² + 9t+ 3 Its direction of motion changes when

(A) t= 1 only

(B) t= 2 only.

(C) t= 3 only

(D) t= 1 and 3

(E) t= 1, 2, and 3

Answer 1

The direction of motion changes when the velocity changes sign.

s(t) = t³ - 6t² + 9t + 3

velocity = s'(t) = 3t² - 12t + 9 = 3(t² - 4t + 3) = 3(t-3)(t-1) = 0 when t = 1 or 3

When 0<t<1, s'(t) > 0

When 1<t<3, s'(t) < 0

When t>3, s'(t) > 0

s'(t) changes sign when t = 1 and when t = 3, so the particle changes direction at those two values of t.

Answer 2

take the derivative and set = 0, solve for t

3t^2 -12t +9 = 0

(3t-3)(t-3) = 0

3t-3 = 0 t-3 = 0

t =3/3 =1 t = 3

It changes direction at 1 and 3

It changes direction, potentially one time less than the highest degree: 3-1 = 2 changes of direction