Finding when the particle is at rest

Differentiate $s(t) = 2t^3 - 15t^2 + 24t + 10$ to get the velocity function $v(t) = s'(t) = 6t^2 - 30t + 24$.

Set $v(t) = 0$: $6t^2 - 30t + 24 = 0$, which simplifies to $t^2 - 5t + 4 = 0$, or $(t-1)(t-4) = 0$. This gives $t = 1$ and $t = 4$, both within $[0,5]$.

Using $v(t) = 6t^2 - 30t + 24$, we find $v(2) = 6(4) - 30(2) + 24 = 24 - 60 + 24 = -12$, so the particle is moving in the negative direction at $t=2$.

The acceleration is $a(t) = v'(t) = 12t - 30$, so $a(2) = 24 - 30 = -6$. Since $v(2) = -12$ and $a(2) = -6$ have the same sign (both negative), the particle is speeding up at $t = 2$.

Evaluate the position at the critical times and endpoints: $s(0) = 10$, $s(1) = 2 - 15 + 24 + 10 = 21$, $s(4) = 2(64) - 15(16) + 24(4) + 10 = 128 - 240 + 96 + 10 = -6$, and $s(5) = 2(125) - 15(25) + 24(5) + 10 = 250 - 375 + 120 + 10 = 5$.

Compute the distance on each subinterval: from $t=0$ to $1$, $|21-10| = 11$; from $t=1$ to $4$, $|-6-21| = 27$; from $t=4$ to $5$, $|5-(-6)| = 11$. Summing these gives the total distance traveled: $11 + 27 + 11 = 49$ meters.

To determine the direction of motion between $t=1$ and $t=4$, we evaluate the velocity at a test point in this interval. At $t=2$, we have $v(2) = 6(4) - 30(2) + 24 = -12 < 0$.

Since velocity is negative at $t=2$ and the velocity function is continuous with no zeros in the open interval $(1,4)$, the velocity remains negative throughout the interval $[1,4]$.

Therefore, the particle is moving backward, that is, in the negative direction between $t=1$ and $t=4$. This matches the earlier distance computation, where the position dropped from 21 meters down to $-6$ meters between these times.