Question

A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 3.8 m/s. The car is a distance d away. The bear is 22 m behind the tourist and running at 6.3 m/s. The tourist reaches the car safely. What is the maximum possible value for d?

Answer 1

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,

`t=(d)/(v_2) =(d+22)/(v_1)`

Substitute 3.8m/s for v₂ and 6.3 m/s for v₁. Solve for d.

`(d)/(v_2) =(d+22)/(v_1)\\ (d)/(3.8 m/s) =(d+22)/(6.3 m/s)\\ 6.3d=3.8d+83.6\\ d=(83.6)/(2.5) \\ d=33.4 m`

The maximum distance between the tourist and his car, for him to be safe from the chasing bear is 33.4 m.