Interval where velocity is increasing

Differentiate $s(t) = t^4 - 4t^3$ to get $v(t) = 4t^3 - 12t^2$, and differentiate again to get $a(t) = 12t^2 - 24t = 12t(t-2)$. Setting $a(t) = 0$ gives $t = 0$ and $t = 2$, which split $[0,4]$ into the subintervals $(0,2)$ and $(2,4)$.

Testing a point in $(0,2)$, say $t=1$: $a(1) = 12(1)(1-2) = -12 < 0$, so velocity is decreasing there. Testing a point in $(2,4)$, say $t=3$: $a(3) = 12(3)(3-2) = 36 > 0$, so velocity is increasing there. Thus velocity increases on $(2,4)$.

From part (a), $v(t) = 4t^3 - 12t^2 = 4t^2(t-3)$, so $v(t) = 0$ at $t=0$ and $t=3$. Testing $t=1$ in $(0,3)$: $v(1) = 4(1)^3 - 12(1)^2 = 4 - 12 = -8 < 0$. Testing $t=3.5$ in $(3,4)$: $v(3.5) = 4(3.5)^3 - 12(3.5)^2 = 171.5 - 147 = 24.5 > 0$. Thus the particle moves in the negative direction on $(0,3)$ and in the positive direction on $(3,4)$, reversing at $t=3$.

Evaluate positions: $s(0) = 0$, $s(3) = 81 - 108 = -27$, $s(4) = 256 - 256 = 0$. The distance on $[0,3]$ is $|s(3)-s(0)| = |-27-0| = 27$, and the distance on $[3,4]$ is $|s(4)-s(3)| = |0-(-27)| = 27$.

Adding these pieces gives the total distance traveled: $27 + 27 = 54$ meters.

From part (b), velocity is negative on (0,3) and positive on (3,4), so the particle reverses direction at t=3. Just after t=3, velocity becomes positive, meaning the particle moves in the positive direction. This matches the position increasing back from s(3)=-27 toward s(4)=0.