The question involves finding the position function $s(t)$ of a particle given the acceleration $a(t) = t^2 - 9t + 4$, along with the initial conditions $s(0) = 0$ and $s(1) = 20$.

Here’s how we solve this step-by-step:

1. Integrate $a(t)$ to find $v(t)$ (velocity function):

$v(t) = \int a(t) \, dt = \int (t^2 - 9t + 4) \, dt$

$v(t) = \frac{t^3}{3} - \frac{9t^2}{2} + 4t + C_1$ where $C_1$ is the integration constant.

2. Integrate $v(t)$ to find $s(t)$ (position function):

$s(t) = \int v(t) \, dt = \int \left(\frac{t^3}{3} - \frac{9t^2}{2} + 4t + C_1 \right) \, dt$

$s(t) = \frac{t^4}{12} - \frac{3t^3}{2} + 2t^2 + C_1t + C_2$ where $C_2$ is another integration constant.

3. Use initial conditions to find $C_1$ and $C_2$:

Condition 1: $s(0) = 0$

Substitute $t = 0$ into $s(t)$: $s(0) = \frac{(0)^4}{12} - \frac{3(0)^3}{2} + 2(0)^2 + C_1(0) + C_2 = 0$ $C_2 = 0$

Condition 2: $s(1) = 20$

Substitute $t = 1$ into $s(t)$: $s(1) = \frac{(1)^4}{12} - \frac{3(1)^3}{2} + 2(1)^2 + C_1(1) + 0 = 20$ $\frac{1}{12} - \frac{3}{2} + 2 + C_1 = 20$ Simplify: $\frac{1}{12} - \frac{18}{12} + \frac{24}{12} + C_1 = 20$ $\frac{7}{12} + C_1 = 20$ $C_1 = 20 - \frac{7}{12} = \frac{240}{12} - \frac{7}{12} = \frac{233}{12}$

4. Final position function $s(t)$:

Substitute $C_1 = \frac{233}{12}$ and $C_2 = 0$ into $s(t)$: $s(t) = \frac{t^4}{12} - \frac{3t^3}{2} + 2t^2 + \frac{233}{12}t$

Let me know if you'd like more details or clarification.

Questions for Further Exploration:

1. How do initial conditions affect the integration constants?
2. Can you derive velocity from position? If so, how?
3. What would happen if one of the conditions, $s(0)$ or $s(1)$, were not provided?
4. Can you interpret the physical significance of $C_1$ and $C_2$?
5. How does changing $a(t)$ affect $s(t)$?

Tip: Always verify units when working with kinematics problems; acceleration, velocity, and position should align dimensionally.