Question

A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.2 m/s. The car is a distance d away. The bear is 29 m behind the tourist and running at 6.0 m/s. The tourist reaches the car safely. What is the maximum possible value for d

Answer 1

Explanation:

It is known that the relation between speed and distance is as follows.

               velocity = $\frac{distance}{time}$

As it is given that velocity is 6 m/s and distance traveled by the bear is (d + 29). Therefore, time taken by the bear is calculated as follows.

         $t_{bear} = \frac{(d + 29)}{6 m/s}$ ............. (1)

As the tourist is running in a car at a velocity of 4.2 m/s. Hence, time taken by the tourist is as follows.

              $t_{tourist} = \frac{d}{4.2}$ ............. (2)

Now, equation both equations (1) and (2) equal to each other we will calculate the value of d as follows.

              $t_{bear} = t_{tourist}$

       $\frac{(d + 29)}{6 m/s}$ = $\frac{d}{4.2}$

                   4.2d + 121.8 = 6d

                         d = $\frac{121.8}{1.8}$

                            = 67.66

Thus, we can conclude that the maximum possible value for d is 67.66.