Answer
(a) The particle reaches minimum velocity at t = 1 s
The minimum velocity is -6 m/s
(b) The acceleration is zero at t = 1 s
Work Step by Step
(a) $x(t) = 2t^3-6t^2+12$
$v(t) = \frac{dx}{dt} = 6t^2-12t$
$v(t) = 6(t-1)^2-6$
The velocity versus time graph is a parabola with a vertex at (1,-6).
This means that the particle reaches minimum velocity at t = 1 s and that
the minimum velocity is -6 m/s.
(b) $a(t) = \frac{dv}{dt} = 12t-12$
$12t-12 = 0$
$t = 1~s$
The acceleration is zero at t = 1 s.
