Answer
$$\eqalign{
& \left( {\text{a}} \right){\text{Graph}} \cr
& \left( {\text{b}} \right)Yes,{\text{ the speed limit is 100}} \cr} $$
Work Step by Step
\[\begin{gathered}
{\text{We have the following table:}} \hfill \\
\boxed{\begin{array}{*{20}{c}}
t&5&{10}&{15}&{20}&{25}&{30} \\
S&{28}&{56}&{79}&{90}&{93}&{94}
\end{array}} \hfill \\
\end{gathered} \]
$$\eqalign{
& {\text{The model data for }}S{\text{ is given by }}S = \frac{{100{t^2}}}{{65 + {t^2}}},{\text{ }}t > 0 \cr
& \cr
& \left( {\text{a}} \right){\text{Using a graphing utility to plot }}S = \frac{{100{t^2}}}{{65 + {t^2}}} \cr
& \left( {{\text{See graph below}}} \right) \cr
& \cr
& \left( {\text{b}} \right){\text{We can notice on the table that when }}t{\text{ increases }}S \cr
& {\text{also increases, but slower}}{\text{. We can calculate the exact}} \cr
& {\text{value evaluating the limit when }}t \to \infty \cr
& \mathop {\lim }\limits_{t \to \infty } S = \mathop {\lim }\limits_{t \to \infty } \frac{{100{t^2}}}{{65 + {t^2}}} \cr
& = \mathop {\lim }\limits_{t \to \infty } \frac{{\frac{{100{t^2}}}{{{t^2}}}}}{{\frac{{65}}{{{t^2}}} + \frac{{{t^2}}}{{{t^2}}}}} = \mathop {\lim }\limits_{t \to \infty } \frac{{100}}{{\frac{{65}}{{{t^2}}} + 1}} \cr
& {\text{Evaluating the limit}} \cr
& \mathop {\lim }\limits_{t \to \infty } \frac{{100}}{{\frac{{65}}{{{t^2}}} + 1}} = \frac{{100}}{{\frac{{65}}{\infty } + 1}} = \frac{{100}}{{0 + 1}} = 100 \cr
& {\text{The speed limit is 100}} \cr} $$
