Homework Help: physics

Posted by Shundale on Sunday, September 7, 2008 at 6:11pm.

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?

physics - DrBob222, Sunday, September 7, 2008 at 6:22pm
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.
physics - Damon, Sunday, September 7, 2008 at 6:29pm
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

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