Answer
(a) $p\left( s \right)=-\frac{1}{150}{{s}^{2}}-\frac{1}{10}s+\frac{68}{3}$.
(b) $1\ \text{ft}$.
Work Step by Step
(a)
The function in standard form is written as $p\left( t \right)=a{{t}^{2}}+bt+c$,where t denotes the time.
So, the equations will be:
Put $t=10$in the equation $p\left( t \right)=a{{t}^{2}}+bt+c$.
$a{{\left( 10 \right)}^{2}}+10b+c=21$ …… (1)
Put $t=20$ in the equation $p\left( t \right)=a{{t}^{2}}+bt+c$. …… (2)
Put $t=40$in the equation $p\left( t \right)=a{{t}^{2}}+bt+c$. …… (3)
Now, subtract equation 2 from equation 1:
$\begin{align}
& a{{\left( 10 \right)}^{2}}+b\left( 10 \right)+c-21-\left( a{{\left( 20 \right)}^{2}}+b\left( 20 \right)+c-18 \right)=0 \\
& a{{\left( 10 \right)}^{2}}+b\left( 10 \right)+c-21-a{{\left( 20 \right)}^{2}}-b\left( 20 \right)-c+18=0 \\
& a\left( 100-400 \right)+b\left( 10-20 \right)=3
\end{align}$
On further simplification:
$\begin{align}
& a\left( -300 \right)+b\left( -10 \right)=3 \\
& -300a-10b=3
\end{align}$
Now, …… (4)
$-300a-10b=3$
Now, subtract equation 2 from equation 3:
$\begin{align}
& a{{\left( 40 \right)}^{2}}+b\left( 40 \right)+c-8-\left( a{{\left( 20 \right)}^{2}}+b\left( 20 \right)+c-18 \right)=0 \\
& a{{\left( 40 \right)}^{2}}+b\left( 40 \right)+c-8-a{{\left( 20 \right)}^{2}}-b\left( 20 \right)-c+18=0 \\
& a\left( 1600-400 \right)+b\left( 40-20 \right)=-10 \\
& 1200a+20b=-10
\end{align}$
Now,
$1200a+20b=-10$ …… (5)
Now multiply equation 4 by 4
$\begin{align}
& 4\left( -300a-10b-3 \right)=0 \\
& -1200a-40b-12=0 \\
\end{align}$
Now, add the above equation to equation 5:
$\begin{align}
& 1200a+20b+10+\left( -1200a-40b-12 \right)=0 \\
& 1200a+20b+10-1200a-40b-12=0 \\
& -20b-2=0 \\
& -20b=2
\end{align}$
Solve further,
$\begin{align}
& -20b=2 \\
& b=-\frac{2}{20} \\
& =-\frac{1}{10}
\end{align}$
Now, substituting the value of b in any of the above:
$\begin{align}
& -300a-10b=3 \\
& -300a-10\cdot \left( -\frac{1}{10} \right)=3 \\
& -300a+1=3 \\
& -300a=2
\end{align}$
Solve further,
$\begin{align}
& a=-\frac{2}{300} \\
& =-\frac{1}{150}
\end{align}$
Now, substitute the values of a and b in equation 1, to get the value of c:
$\begin{align}
& \left( -\frac{1}{150} \right){{\left( 10 \right)}^{2}}+\left( -\frac{1}{10} \right)\left( 10 \right)+c=21 \\
& \left( -\frac{1}{150} \right)\cdot 100+\left( -\frac{1}{10} \right)\cdot 10+c=21 \\
& -\frac{2}{3}-1+c=21 \\
& -\frac{5}{3}+c=21
\end{align}$
On further simplification:
$\begin{align}
& c=21+\frac{5}{3} \\
& =\frac{68}{3}
\end{align}$
Thus, the quadratic equation of the given data will be written as: $p\left( s \right)=-\frac{1}{150}{{s}^{2}}-\frac{1}{10}s+\frac{68}{3}$.
(b)
$p\left( s \right)=-\frac{1}{150}{{s}^{2}}-\frac{1}{10}s+\frac{68}{3}$
Now, take $s=50$and substitute in the quadratic equation,
So,
$\begin{align}
& p\left( 50 \right)=-\frac{1}{150}{{s}^{2}}-\frac{1}{10}s+\frac{68}{3} \\
& =-\frac{1}{150}{{\left( 50 \right)}^{2}}-\frac{1}{10}\left( 50 \right)+\frac{68}{3} \\
& =-\frac{50}{3}-5+\frac{68}{3} \\
& =-5+6
\end{align}$
Therefore, the lift distance for a flow rate of 50 gal per min is 1 ft.
