Let v(t)=3t^2-12t be the velocity of a particle moving along the x-axis for time t, t is greater then or equal to 0. When t=0, the particle is at x=-6

Question

A: determine the position of the particle at t=a

B: write an expression for the speed of the particle at any time t.

C: In which direction does the particle begin moving and when does it turn around?

D: When, if ever, is the particle at the origin?

Answer 1

A) Integrate v(t): x = -6 + ∫₀^a (3t² -12t) dt = -6 + [t³ - 6t²]₀^a = a³ - 6a² - 6.

B) Speed is magnitude of v, so speed = |3t² - 12t|.

C) Displacement x decreases at first, going to more negative numbers; the turn is at v = 0 = 3t(t-4), t = 4 sec.

D) With x = a³ - 6a² - 6 = a²(a - 6) - 6 going to zero, we must have: a > 6. I find that x becomes positive at t = 6.158 sec (don't have graphing calculator but know old-fashioned method of iteration, guessing different values and seeing how the results trend).