The maximum possible value for d is 33.4 m.
The bear chases the tourist with a speed v₁ and the tourist runs with a speed v₂. The distance between the tourist and the car is d and the distance between the tourist and the bear is 22 m. for the tourist to reach the car safely, he should cover the distance d in the same time as the bear, which covers a distance d+22 m.
Therefore,
[tex]t=\frac{d}{v_2} =\frac{d+22}{v_1}[/tex]
Substitute 3.8m/s for v₂ and 6.3 m/s for v₁. Solve for d.
[tex]\frac{d}{v_2} =\frac{d+22}{v_1}\\ \frac{d}{3.8 m/s} =\frac{d+22}{6.3 m/s}\\ 6.3d=3.8d+83.6\\ d=\frac{83.6}{2.5} \\ d=33.4 m[/tex]
The maximum distance between the tourist and his car, for him to be safe from the chasing bear is 33.4 m.
