Answer
(a) $p\left( s \right)=\left( \frac{3}{16} \right){{s}^{2}}-\frac{135}{4}s+1750$.
(b) 531
Work Step by Step
(a)
The function in standard form is written as $p\left( s \right)=a{{s}^{2}}+bs+c$, where s denotes the travel speed.
So, the equations will be:
Put $s=60$ in the equation $p\left( s \right)=a{{s}^{2}}+bs+c$.
$a{{\left( 60 \right)}^{2}}+b\left( 60 \right)+c=400$ …… (1)
Put $s=80$ in the equation $p\left( s \right)=a{{s}^{2}}+bs+c$.
$a{{\left( 80 \right)}^{2}}+b\left( 80 \right)+c=250$ …… (2)
Put $s=100$in the equation $p\left( s \right)=a{{s}^{2}}+bs+c$.
$a{{\left( 100 \right)}^{2}}+b\left( 100 \right)+c=250$ …… (3)
Now, subtract equation (2) from equation (1):
$\begin{align}
& a{{\left( 60 \right)}^{2}}+b\left( 60 \right)+c-400-\left( a{{\left( 80 \right)}^{2}}+b\left( 80 \right)+c-250 \right)=0 \\
& a{{\left( 60 \right)}^{2}}+b\left( 60 \right)+c-400-a{{\left( 80 \right)}^{2}}-b\left( 80 \right)-c+250=0 \\
& a\left( 3600-6400 \right)-b\left( 60-80 \right)-150=0 \\
& -2800a-20b-150=0
\end{align}$
Now,
$-2800a-20b-150=0$ …… (4)
Now, subtract equation (2) from equation (3):
$\begin{align}
& a{{\left( 100 \right)}^{2}}+b\left( 100 \right)+c-250-\left( a{{\left( 80 \right)}^{2}}+b\left( 80 \right)+c-250 \right)=0 \\
& a{{\left( 100 \right)}^{2}}+b\left( 100 \right)+c-250-a{{\left( 80 \right)}^{2}}-b\left( 80 \right)-c+250=0 \\
& a\left( 10000-6400 \right)+b\left( 100-80 \right)=0 \\
& 3600a+20b=0
\end{align}$
Now,
$3600a+20b=0$ …… (5)
Now add equation (4) and equation (5):
$\begin{align}
& 3600a+20b+\left( -2800a-20b-150 \right)=0 \\
& 3600a+20b-2800a-20b-150=0 \\
& 800a-150=0
\end{align}$
Upon further simplification:
$\begin{align}
& 800a=150 \\
& a=\frac{150}{800} \\
& =\frac{3}{16}
\end{align}$
Now, substituting the value of a in any of the above equations:
$\begin{align}
& -2800\left( \frac{3}{16} \right)-20b=150 \\
& -525-20b=150 \\
& -20b=675
\end{align}$
Divide by $-20$ on both sides
$\begin{align}
& b=\frac{675}{-20} \\
& =-\frac{135}{4}
\end{align}$
Now, substitute the values of a and b in equation 1, to get the value of c:
$\begin{align}
& a{{\left( 60 \right)}^{2}}+b\left( 60 \right)+c=400 \\
& \left( \frac{3}{16} \right){{\left( 60 \right)}^{2}}-\left( \frac{135}{4} \right)\left( 60 \right)+c=400 \\
& \left( \frac{3}{16} \right)\cdot 3600-\left( \frac{135}{4} \right)\left( 60 \right)+c=400 \\
& 675-2025+c=400
\end{align}$
Upon further simplification
$c=1750$
Thus, the quadratic equation of the given data will be written as: $p\left( s \right)=\left( \frac{3}{16} \right){{s}^{2}}-\frac{135}{4}s+1750$
(b)
$p\left( s \right)=\left( \frac{3}{16} \right){{t}^{2}}-\frac{135}{4}t+1750$
Now, take $s=50$ and substitute in the quadratic equation:
So,
$\begin{align}
& p\left( 50 \right)=\frac{3}{16}\cdot {{\left( 50 \right)}^{2}}-\frac{135}{4}\cdot \left( 50 \right)+1750 \\
& =\frac{3}{16}\cdot \left( 2500 \right)-\frac{135}{4}\cdot \left( 50 \right)+1750 \\
& =468.75-1687.5+1750 \\
& =531.25
\end{align}$
Therefore, the number of accidents that are estimated that would occur at $50\ \text{km/hr}$ are 531 accidents.
