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 27 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?
distance = rate x time and rearrange to
time = distance/rate.
time for bear = d+27/6
time for tourist = d/3.8
The time is the same so set the time equal to each other and solve for distance. Post your work if you get stuck.
both take the same amount of time, t
distance of tourist from car d = 3.8 t
distance of bear from car = 3.8 t + 27 = 6 t
so
2.2 t = 27
t = 270/22= about 12.3 seconds
d = 3.8 t = 3.8 * 270/22 = 46.6 m max
To find the maximum possible value for d, let's break down the scenario and calculate the time it takes for the tourist to reach the car.
The tourist is running with a speed of 3.8 m/s and the bear is running at 6.0 m/s. So, the relative speed between the bear and the tourist is the difference of their speeds, which is 6.0 m/s - 3.8 m/s = 2.2 m/s.
The bear is initially 27 m behind the tourist. To catch up to the tourist, it needs to reduce this distance by moving at a relative speed of 2.2 m/s.
Time taken for the bear to reduce the distance by 27 m is given by:
Time = Distance / Speed
Time = 27 m / 2.2 m/s
Time ≈ 12.27 seconds
Now, let's calculate the time it takes for the tourist to reach the car:
The distance between the tourist and the car, d, is constant and in this case, equal to the distance that the bear needs to cover, which is 27 m.
Time taken for the tourist to reach the car is given by:
Time = Distance / Speed
Time = 27 m / 3.8 m/s
Time ≈ 7.11 seconds
Since the tourist reaches the car safely, the time taken for the tourist to reach the car must be less than the time taken for the bear to reduce the distance by 27 m.
Therefore, d ≤ 27 m.
Thus, the maximum possible value for d is 27 meters.