Jackie Chan and Matt Damon star in “Some Action Movie.” They realize that a parked car is about to explode. At that point, both Jackie and Matt are standing 2.00 m from the car. Jackie has a faster reaction time and immediately begins running away from th

a) Jackie's position function: Jackie's initial position is 2.00 m from the car. She begins running away from the car at a constant speed of 5.80 m/s. So her position function can be written as: x_jackie = 2.00 m + 5.80 m/s t b) Matt's position function: Matt starts running 0.90 s later than Jackie, so his position function will start at a later time. Matt starts running at time t = 0.90 s, so his position function can be written as: x_matt = 6.70 m/s (t - 0.90 s) c) How long does Matt run before catching up to Jackie? Let's call the time at which Matt catches up to Jackie as t_catch. At that time, their positions will be equal: 2.00 m + 5.80 m/s t_catch = 6.70 m/s (t_catch - 0.90 s) Solving for t_catch, we get: t_catch = 3.56 s d) When Matt catches up to Jackie, how far from the car are they? At time t_catch, both Jackie and Matt have run different distances from the car. We can use either of their position functions to find their positions at time t_catch, and then find the distance between them. Let's use Jackie's position function: x_jackie(t_catch) = 2.00 m + 5.80 m/s (3.56 s) = 22.48 m At time t_catch, Matt's position is: x_matt(t_catch) = 6.70 m/s (3.56 s - 0.90 s) = 31.64 m So when Matt catches up to Jackie, they are 31.64 m - 22.48 m = 9.16 m away from the car.