Answer
(a) $x'(t)$ is velocity and $x''(t)$ is acceleration.
$x'(t) = 3t^2 - 12t + 9$
$x''(t) = 6t - 12$
(b) $(0, 1)$ U $(3, 5)$
(c) $-3$
Work Step by Step
(a) $x(t) = t^3 - 6t^2 + 9t - 2$
To find the velocity function, differentiate the position function.
$x'(t) = 3t^2 - 12t + 9$
To find the acceleration function, differentiate the velocity function.
$x''(t) = 6t - 12$
(b) The particle moves to the right when the velocity is positive. Therefore, we must find the intervals on which $x'(t) > 0$.
Start by finding where the velocity is zero.
$0 = 3t^2 - 12t + 9$
$0 = 3(t^2 - 4t + 3)$
$0 = 3(t - 3)(t - 1)$
$v(t) = 0$ at $t = 1$ and $t = 3$
Find if the velocity is positive or negative on each interval by substituting a time from each interval.
$t < 1$:
$x'(0) = 3(0 - 3)(0 - 1) > 0$
$1 < t < 3$:
$x'(2) = 3(2 - 3)(2 - 1) < 0$
$t > 3$:
$x'(4) = 3(4 - 3)(4 - 1) > 0$
So, velocity is positive when $t < 1$ and when $t > 3$. Since the position function only applies to the time interval $0 < t < 5$, the intervals on which the particle is moving right are:
$(0, 1)$ and $(3, 5)$
(c) Find the time at which acceleration is zero by setting $x''(t) = 0$.
$0 = 6t - 12$
$6t = 12$
$t = 2$
Now, find the velocity at this time.
$x''(2) = 3(2)^2 - 12(2) + 9$
$x''(2) = 3(4) - 24 + 9$
$x''(2) = 12 - 24 + 9$
$x''(2) = -3$
