• anonymous
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 30 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?
  • Stacey Warren - Expert
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
  • katieb
I got my questions answered at in under 10 minutes. Go to now for free help!
  • kropot72
Time taken for the tourist to reach the car is \[t=\frac{distance}{speed}=\frac{d}{4.2}\ ........(1)\] Time taken for the tourist to reach the car is \[t=\frac{distance}{speed}=\frac{d+30}{6}\ ......(2)\] If the tourist and the bear reach the car at the same time, we can equate (1) and (2) giving: \[\frac{d}{4.2}=\frac{d+30}{6}\ ...........(3)\] Cross multiplying and simplifying equation (3) gives 1.8d = 126 which leads to a maximum possible value for d of: \[d=\frac{126}{1.8}=you\ can\ calculate\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.