Answer
From $t = 0.0~s$ to $t = 4.0~s$, the particle moves a distance of 82 meters.
Work Step by Step
$x = 3.0t^2-2.0t^3$
$v = 6.0t-6.0t^2$
On the interval $(0,1)$, the velocity is positive.
On the interval $(1,4)$, the velocity is negative.
At $t = 0,$ $~~~x = 3.0(0)^2-2.0(0)^3 = 0~m$
At $t = 1,$ $~~~x = 3.0(1)^2-2.0(1)^3 = 1~m$
At $t = 4,$ $~~~x = 3.0(4)^2-2.0(4)^3 = -80~m$
From $t = 0.0~s$ to $t = 1.0~s$, the particle moves a distance of 1.0 meter from $x = 0.0~m$ to $x = 1.0 m$
From $t = 1.0~s$ to $t = 4.0~s$, the particle moves a distance of 81 meters from $x = 1.0~m$ to $x = -80 m$
From $t = 0.0~s$ to $t = 4.0~s$, the particle moves a total distance of 82 meters.
