A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.09 m/s. The car is a distance d away. The bear is 20.9 m behind the tourist and running at 6.04 m/s. The tourist reaches the car safely. What is the maximum possible value for d?
I don't know where to start or what to do
t = d/4.09
and
t = (20.9 + d)/6.04
so
d/4.09 = (20.9 + d)/6.04
d + 20.9 = 1.48 d
.48 d = 20.9
d = 20.9/.48 meters
To find the maximum possible value for d, we need to determine the condition where the bear is just at the point of catching up to the tourist as the tourist reaches their car. At this point, the distance traveled by the bear and the distance traveled by the tourist should be equal.
Let's analyze the situation step by step:
1. The relative distance between the tourist and the bear initially is 20.9 meters.
2. The tourist's speed is 4.09 m/s, and the bear's speed is 6.04 m/s.
3. The bear is chasing the tourist, so its relative speed compared to the tourist is the difference between their speeds: 6.04 m/s - 4.09 m/s = 1.95 m/s.
Now, we can use the formula:
Relative distance = Relative speed × Time
Let's denote the time taken to reach the car as T.
For the bear, the relative distance covered is:
Relative distance = Relative speed × Time
Relative distance = 1.95 m/s × T
For the tourist, the distance covered is:
Distance = Speed × Time
Distance = 4.09 m/s × T
At the moment the tourist reaches the car, the relative distances covered by the tourist and the bear should be equal. Therefore, we can set up an equation:
4.09 m/s × T = 1.95 m/s × T + 20.9 m
Simplifying the equation, we have:
4.09 m/s × T - 1.95 m/s × T = 20.9 m
2.14 m/s × T = 20.9 m
T = 20.9 m / 2.14 m/s
T ≈ 9.77 seconds
Now, to find the maximum possible value for d, we use the equation for distance covered by the tourist:
Distance = Speed × Time
d = 4.09 m/s × 9.77 s
d ≈ 39.97 m
Therefore, the maximum possible value for d is approximately 39.97 meters.