A particle is moving with the given data. Find the position of the particle, $s(t)$.
\[
a(t)=t^{2}-5 t+3, \quad s(0)=0, \quad s(1)=20
\]
\[
s(t)=
\]

Step 1 ：\(a(t) = t^2 - 5t + 3\)

Step 2 ：\(v(t) = \int a(t) dt = \frac{1}{3}t^3 - \frac{5}{2}t^2 + 3t + C_1\)

Step 3 ：\(v(0) = 0 = \frac{1}{3}(0)^3 - \frac{5}{2}(0)^2 + 3(0) + C_1 \Rightarrow C_1 = 0\)

Step 4 ：\(v(t) = \frac{1}{3}t^3 - \frac{5}{2}t^2 + 3t\)

Step 5 ：\(s(t) = \int v(t) dt = \frac{1}{12}t^4 - \frac{5}{6}t^3 + \frac{3}{2}t^2 + C_2\)

Step 6 ：\(s(0) = 0 = \frac{1}{12}(0)^4 - \frac{5}{6}(0)^3 + \frac{3}{2}(0)^2 + C_2 \Rightarrow C_2 = 0\)

Step 7 ：\(s(1) = 20 = \frac{1}{12}(1)^4 - \frac{5}{6}(1)^3 + \frac{3}{2}(1)^2\)

Step 8 ：\(20 = \frac{1}{12} - \frac{5}{6} + \frac{3}{2}\)

Step 9 ：\(\frac{1}{12} - \frac{5}{6} + \frac{3}{2} = \frac{1 - 10 + 9}{6} = \frac{0}{6} = 0\)

Step 10 ：\(C_2 = 20\)

Step 11 ：\(s(t) = \frac{1}{12}t^4 - \frac{5}{6}t^3 + \frac{3}{2}t^2 + 20\)

Step 12 ：\(\boxed{s(t) = \frac{1}{12}t^4 - \frac{5}{6}t^3 + \frac{3}{2}t^2 + 20}\)