A car is speed at a constant 24 m/sec in a school zone. A police car starts from rest just as the speeder passes it and accelerates at a constant rate of 5.4 m/sec2. Also, the cop catches the car at t=8.89 s and v=48. How fast is the police car going when it is 27 m behind the speeding car for the second time? So, I'm not sure if the wording here is what's confusing me, but I can't get the right answer. First of all, how can the cop car ever be behind the speeding car "for the second time"? If the car is at a constant velocity and the police car keeps accelerating, then once they meet for the first time, the cop car will always be ahead, so what does it mean for it to be behind the speeding car for the second time? I attempted the problem using basic kinematic equations to try to first find when it is 27 m behind the car: v=48+5.4t and 27=0+48t+.5(5.4)t^2. And so I solved for t and plugged it into the first equation but that didn't get me the right answer. Maybe I'm just not understanding the problem..