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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?

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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