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 26 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?

d2 = d1 + 26 m.

r2*t = r1*t + 26
6t = 4.2t + 26
1.8t = 26
t = 14.44 s.

d = d1 = 4.2*14.44 = 60.67 m..

To find the maximum possible value for d, we need to find the time it takes for the tourist to reach the car and compare it to the time it takes for the bear to catch up to the tourist.

Let's assume the time it takes for the tourist to reach the car is t. To find t, we can use the equation:

Distance = Speed × Time

The distance the tourist runs is d. Therefore, we have:

d = 4.2 m/s × t

Now let's find the time it takes for the bear to catch up to the tourist. Since the bear starts 26 m behind the tourist and they are both traveling in the same direction, the distance between them is decreasing at a rate of (6.0 m/s - 4.2 m/s) = 1.8 m/s.

The time it takes for the bear to catch up to the tourist is given by:

Time = distance / relative speed

Since the distance between them is 26 m and the relative speed is 1.8 m/s, we have:

Time = 26 m / 1.8 m/s ≈ 14.4 s

Now we can compare the time it takes for the tourist to reach the car (t) to the time it takes for the bear to catch up (14.4 s). The tourist must reach the car before the bear catches up, so:

t < 14.4 s

Now we can solve for the maximum possible value of d:

d = 4.2 m/s × t
d < 4.2 m/s × 14.4 s
d < 60.48 m

Therefore, the maximum possible value for d is approximately 60.48 meters.