1. The problem statement, all variables and given/known data A speeder zooms past a parked police car at a constant speed of 70 mph. Then, 3 sec later, the policewoman starts accelerating from rest at 10 ft/s^2 until her velocity is 85 mph. How long does it take her to overtake the speeding car if it neither slows down nor speeds up? 2. Relevant equations x''=a x'=at+c1 x=(a/2)t^2+c1t+c2 3. The attempt at a solution Speeder velocity = 102.7 ft/s Police velocity = 124/7 ft/s Police equations of motion while accelerating: x''=10 x'=10t x=5t^2 Thus 124.7=10t ---> t=12.47 s Since police starts moving 3 secs later, after 15.47s: Speeder has covered 15.47 * 102.7 = 1588.8 ft Police has covered 5(12.47)^2 = 777.5 ft The distance between them is 811.3 ft. From this point on both are constant velocity: Police: x'=124.7 x=124.7t Speeder: x'=102.7 x=102.7t+811.3 Equating the two equations and solving get t=36s. Book says 52.6 sec. No idea where i went wrong.