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