1. The problem statement, all variables and given/known data Large cockroaches can run as fast as 1.50 m/s in short bursts. Suppose you turn on the light in a cheap motel and see one scurrying directly away from you at a constant 1.50 m/s. If you start 0.91 m behind the cockroach with an initial speed of 0.82 m/s toward it, what minimum constant acceleration would you need to catch up with it when it travelled 1.20 m, just short of safety under a counter? 2. Relevant equations x=x0+v0*t+1/2*a*t^2 3. The attempt at a solution I thought that to solve this I musty first find the time it took the cockroach to get to the counter fro 0.91m to 1.2 m. To do this I used the formula x=x0+v0*t+1/2*a*t^2 (where x=1.2, x0=0.91, v0=1.5m/s and, a=0 and t=?) so I got t= (x-x0)/v0 t= 0.29/1.5 =0.193 seconds Then I used x=x0+v0*t+1/2*a*t again and rearranged it to find the acceleration of the man. I inserted the equation with x=1.2, x0=0, v0=0.82, a=??, t=0.193 to get: a= (x-x0- v0*t )/ (1/2*t^2) a= (1.2 - 0.82*0.193) / (0.5 * 0.193^2) a=57.71 m/s2 which is wrong according to the answer :( Please help, use simple physics as I am very new to the subject. And thank you in advance !!