Answer
(a) $v = 3t^2-12$
$a = 6t$
(b) The particle is moving upward when $t \gt 2$
The particle is moving downward when $0 \lt t \lt 2$
(c) 23
(d) We can see the graphs below.
(e) When $t \gt 2$, the particle is speeding up.
When $0 \lt t \lt 2$, the particle is slowing down.
Work Step by Step
(a) $y = t^3-12t+3$
$v = \frac{dy}{dt} = 3t^2-12$
$a = \frac{dv}{dt} = 6t$
(b) We can find the time $t$ when $v = 0$:
$v = 3t^2-12 = 0$
$3(t^2-4) = 0$
$(t-2)(t+2) = 0$
$v=0~~$ when $~~t = 2$
When $t \gt 2$, the velocity is positive.
Thus the particle is moving upward when $t \gt 2$
When $0 \lt t \lt 2$, the velocity is negative.
Thus the particle is moving downward when $0 \lt t \lt 2$
(c) At $t = 0$:
$y = (0)^3-12(0)+3 = 3$
At $t = 2$:
$y = (2)^3-12(2)+3 = -13$
At $t = 3$:
$y = (3)^3-12(3)+3 = -6$
We can find the distance $d$ the particle travels in the first 3 seconds:
$d = 16+7 = 23$
(d) We can see the graphs below.
(e) The particle is speeding up when the magnitude of the velocity is increasing. Thus, the particle is speeding up when the velocity is moving farther from the horizontal axis, and the particle is slowing down when the velocity is moving closer to the horizontal axis,
When $t \gt 2$, the particle is speeding up.
When $0 \lt t \lt 2$, the particle is slowing down.