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Lera25 [3.4K]
3 years ago
5

A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 3.9 m/s. The car is a distanc

e d away. The bear is 23 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?
Physics
1 answer:
sammy [17]3 years ago
6 0
If it takes t seconds to reach the car, then the distance d is 3.9t.

The bear's distance from the tourist's starting point is
6t-23

For maximum d, we set the equations equal to each other:
3.9t=6t-23
\Rightarrow -2.1t=-23
\Rightarrow t=\frac{23}{2.1}
so the distance is
d=3.9(\frac{23}{2.1})\approx42.741\ m
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