Atourist being chased by an angry bear is running in a straight line toward his car at a speed of 5.20 m/s. the car is a distance d away. the bear is 25.8 m behind the tourist and running at 7.38 m/s. the tourist reaches the car safely. what is the maximum possible value for d?

We know that both tourist and bear run towards same direction at diferent speed but without acceleration, then they accomplish the follow formula: v=d/t where v is speed, d is distance and t is time. For the tourist: Vt=d/t=5.2 => t=d/5.2, where d is distance from tourist to car and for bear: Vb=(d+25.8)/t=7.38 because bear is 25.8 metres behind tourist, so d+25.8 is the distance from bear to car. The bear will catch the tourist only if it reach the tourist at the same time or before than the tourist reach the car, that means time should be the same in both equations and we need to solve the second one: 7.38=(d+25.8)/t = (d+25.8)/(d/5.2) => 7.38*(d/5.2)=d+25.8 => 7.38/5.2*d-d=25.8 d=25.8/(7.38/5.2-1) = 61.54 m So then, maximum distance from tourist to car should be 61.54 metres or tourist will be catched by the bear.

Ithink the answer would be a