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Q & A: Who eats lunch first, the preacher or the bear?

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Most recent answer: 09/06/2011
Q:
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?
- Em (age 21)
Denver, CO, USA
A:
Hi Em,
Your problem is a straightforward algebra problem with a single variable and shows how you can construct a solvable equation by using invariants.  The givens are two distances to the car and two velocities.  What is not given are the two times of arrival, but you know that they are equal for the maximum distance away, so this is the invariant.  The relationship among distance, time and velocity is  D = VT. Inverting this gives T = D/V.  You know that TP = D/VP and TB= (D+30)/VB.  Setting these two equal you can easily solve for D.   You can do the arithmetic.

The reason for my title is that your scenario is so much like that in a novelty song written the the early 1900's "The Preacher and the Bear".  My favorite version was performed by Phil Harris; you can find it on You-tube. 

LeeH


(published on 09/06/2011)

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