A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 5.48 m/s. The car is a distance d away. The bear is 30.5 m behind the tourist and running at 6.67 m/s. The tourist reaches the car safely. What is the maximum possible value for d?
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