Answer
(a) $p\left( s \right)=\left( 0.05 \right){{s}^{2}}-\left( 5.5 \right)s+250$
(b) 100
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 $p\left( s \right)=a{{s}^{2}}+bs+c$
$a{{\left( 60 \right)}^{2}}+60b+c=100$ …… (1)
Put $s=80$ in the equation $p\left( s \right)=a{{s}^{2}}+bs+c$
$a{{\left( 80 \right)}^{2}}+b\left( 80 \right)+c=130$ …… (2)
Put $s=100$ in the equation $p\left( s \right)=a{{s}^{2}}+bs+c$
$a{{\left( 100 \right)}^{2}}+b\left( 100 \right)+c=200$ …… (3)
Now, subtract equation 2 from equation 1:
$\begin{align}
& a{{\left( 80 \right)}^{2}}+b\left( 80 \right)+c-130-\left( a{{\left( 60 \right)}^{2}}+b\left( 60 \right)+c-100 \right)=0 \\
& a{{\left( 80 \right)}^{2}}+b\left( 80 \right)+c-130-a{{\left( 60 \right)}^{2}}-b\left( 60 \right)-c+100=0 \\
& \left( 6400-3600 \right)a+\left( 80-60 \right)b=30 \\
& 2800a+20b=30
\end{align}$
Now,
$2800a+20b=30$ …… (4)
Now, subtract equation 2 from equation 3:
$\begin{align}
& a{{\left( 100 \right)}^{2}}+b\left( 100 \right)+c-200-\left( a{{\left( 80 \right)}^{2}}+b\left( 80 \right)+c-130 \right)=0 \\
& a{{\left( 100 \right)}^{2}}+b\left( 100 \right)+c-200-a{{\left( 80 \right)}^{2}}-b\left( 80 \right)-c+130=0 \\
& \left( 10000-6400 \right)a+\left( 100-80 \right)b=70 \\
& 3600a+20b=70
\end{align}$
Now,
$3600a+20b=70$ …… (5)
Now subtract equation 4 from equation 5:
$\begin{align}
& 3600a+20b-70-\left( 2800a+20b-30 \right)=0 \\
& 3600a+20b-70-2800a-20b+30=0 \\
& 3600a+20b-2800a-20b=70-30 \\
& 800a=40
\end{align}$
On further simplification:
$800a=40$
Divide by 800 on both sides:
$\begin{align}
& a=\frac{40}{800} \\
& =\frac{5}{100} \\
& =0.05
\end{align}$
Now, substituting the value of a in any of the above equations:
$\begin{align}
& 2800\left( 0.05 \right)+20b=30 \\
& 140+20b=30
\end{align}$
Subtract 140 from both sides:
$\begin{align}
& 140+20b-140=30-140 \\
& 20b=-110
\end{align}$
Divide by 20 on both sides:
$\begin{align}
& b=-\frac{110}{20} \\
& =-5.5
\end{align}$
Now, substitute the values of a and b in equation 1, to get the value of c:
$\begin{align}
& \left( 0.05 \right){{\left( 60 \right)}^{2}}-\left( 5.5 \right)\left( 60 \right)+c=100 \\
& \left( 5 \right)\cdot 36-\left( 55 \right)\cdot 6+c=100 \\
& 180-330+c=100 \\
& c=250
\end{align}$
Thus, the quadratic equation of the given data will be written as: $p\left( s \right)=\left( 0.05 \right){{s}^{2}}-\left( 5.5 \right)s+250$.
(b)
$p\left( s \right)=\left( 0.05 \right){{s}^{2}}-\left( 5.5 \right)s+250$
Now, take $s=50$and substitute in the quadratic equation,
So,
$\begin{align}
& p\left( s \right)=\left( 0.05 \right){{s}^{2}}-\left( 5.5 \right)s+250 \\
& =\left( 0.05 \right){{\left( 50 \right)}^{2}}-\left( 5.5 \right)\left( 50 \right)+250 \\
& =\left( 0.05 \right)\cdot 2500-\left( 5.5 \right)\cdot 50+250 \\
& =125-275+250
\end{align}$
On further simplification
$\begin{align}
& p\left( s \right)=375-275 \\
& =100
\end{align}$
Thus, the number of accidents that are estimated that would occur at $50\ \text{km/hr}$is 100.
