Posted by Shundale on .
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,
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,
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