1. The problem statement, all variables and given/known data Problem: A police car travelling at 80km/h is passed by speeder travelling at 145km/h. 3 seconds after the speeder passes, the police car starts accelerating at 2.5m/s2. How much time passes before the police overtakes the speeder (moving at a constant speed)? Variables: initial velocity of police car (vpi) 80km/h=22.222m/s initial velocity of speeder (vsi)=final velocity of speeder (vsf) = 40.278m/s acceleration of police car (ap)=2.5m/s2 2. Relevant equations x=x0+v0t+1/2*a*t2 3. The attempt at a solution I thought that if the distance the police car travelled equals the distance the speeder travelled, that would mean that the police overtook the speeder. So I used the distance formula/equation like this: time passed since the police started accelerating that the police car and speeder meets=t vsi*3+vsf*t=vpi*3+vpi*t+1/2*a*t2 40.278*3+40.278t = 22.222*3+22.222*t+1/2*2.5*t2 When I did this equation on mathway I got the answers t=13.0062276,1.4385724. So if I add 3 seconds to the answers I get t=16 and t=4. I thought the final answer would be 16 seconds because 4 seconds is too short a time. But it says my answer is incorrect... Could you tell me where I'm thinking wrong? Thank you!