If it takes \(t\) seconds to reach the car, then the distance \(d\) is \(3.9t\).
The bear's distance from the tourist's starting point is
\(6t-23\)
For maximum \(d\), we set the equations equal to each other:
\(3.9t=6t-23\)
\(\Rightarrow -2.1t=-23\)
\(\Rightarrow t=\frac{23}{2.1}\)
so the distance is
\(d=3.9(\frac{23}{2.1})\approx42.741\ m\)
