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

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