1. Problem 1: Given position function $s = t^2 - 3t + 2$ for $0 \leq t \leq 2$. 2. Displacement is the change in position: $\Delta s = s(2) - s(0)$. 3. Average velocity is displacement over time interval: $v_{avg} = \frac{\Delta s}{2 - 0}$. 4. Speed is the magnitude of velocity. Velocity is the first derivative: $v(t) = \frac{ds}{dt} = 2t - 3$. 5. Acceleration is the derivative of velocity: $a(t) = \frac{dv}{dt} = 2$. 6. Evaluate speed and acceleration at endpoints $t=0$ and $t=2$. 7. To find when the body changes direction, solve $v(t) = 0$ within the interval. --- 1. Problem 2: $s = 6t - t^2$, $0 \leq t \leq 6$. 2. Displacement: $s(6) - s(0)$. 3. Average velocity: $\frac{s(6) - s(0)}{6}$. 4. Velocity: $v(t) = 6 - 2t$. 5. Acceleration: $a(t) = -2$. 6. Speed and acceleration at $t=0$ and $t=6$. 7. Change direction when $v(t) = 0$. --- 1. Problem 3: $s = -t^3 + 3t^2 - 3t$, $0 \leq t \leq 3$. 2. Displacement: $s(3) - s(0)$. 3. Average velocity: $\frac{s(3) - s(0)}{3}$. 4. Velocity: $v(t) = -3t^2 + 6t - 3$. 5. Acceleration: $a(t) = -6t + 6$. 6. Speed and acceleration at $t=0$ and $t=3$. 7. Change direction when $v(t) = 0$. --- 1. Problem 4: $s = \frac{t^4}{4} - t^3 + t^2$, $0 \leq t \leq 3$. 2. Displacement: $s(3) - s(0)$. 3. Average velocity: $\frac{s(3) - s(0)}{3}$. 4. Velocity: $v(t) = t^3 - 3t^2 + 2t$. 5. Acceleration: $a(t) = 3t^2 - 6t + 2$. 6. Speed and acceleration at $t=0$ and $t=3$. 7. Change direction when $v(t) = 0$. --- 1. Problem 5: $s = \frac{25}{t^2} - \frac{5}{t}$, $1 \leq t \leq 5$. 2. Displacement: $s(5) - s(1)$. 3. Average velocity: $\frac{s(5) - s(1)}{4}$. 4. Velocity: $v(t) = \frac{d}{dt}\left(25t^{-2} - 5t^{-1}\right) = -50t^{-3} + 5t^{-2}$. 5. Acceleration: $a(t) = \frac{d}{dt}v(t) = 150t^{-4} - 10t^{-3}$. 6. Speed and acceleration at $t=1$ and $t=5$. 7. Change direction when $v(t) = 0$. --- 1. Problem 6: $s = \frac{25}{t+5}$, $-4 \leq t \leq 0$. 2. Displacement: $s(0) - s(-4)$. 3. Average velocity: $\frac{s(0) - s(-4)}{4}$. 4. Velocity: $v(t) = \frac{d}{dt} \left(25(t+5)^{-1}\right) = -25(t+5)^{-2}$. 5. Acceleration: $a(t) = \frac{d}{dt}v(t) = 50(t+5)^{-3}$. 6. Speed and acceleration at $t=-4$ and $t=0$. 7. Change direction when $v(t) = 0$ (check if possible). --- 1. Problem 7: $s = t^3 - 6t^2 + 9t$, position along s-axis. 2. Velocity: $v(t) = 3t^2 - 12t + 9$. 3. Acceleration: $a(t) = 6t - 12$. 4. a) Find acceleration when velocity is zero: solve $v(t) = 0$ for $t$, then find $a(t)$. 5. b) Find speed when acceleration is zero: solve $a(t) = 0$ for $t$, then find $|v(t)|$. 6. c) Total distance from $t=0$ to $t=2$: find positions at critical points where velocity changes sign, sum absolute position changes. --- Final answers are computed by substituting values and simplifying each step as shown above.