Question

An unmarked police car traveling a constant 95km/h is passed by a speeder traveling 133km/h .

Precisely 1.00 s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.50m/s2 , how much time passes after the police car is passed by a speeder and before the police car overtakes the speeder (assumed moving at constant speed)?

Answer 1

We convert the units:

             (95km/h)/(3.6ks/h) = 26.4m/s.

             (133km/h)/(3.6ks/h) = 36.9m/s.

         We use a coordinate system with the origin where the motorist passes the police officer.

         The location of the speeding motorist is given by

              xm= xo + vmt = 0 + (36.9m/s)t.

         The location of the police officer is given by

              xp = xo + vop(1.00s) + vop(t - 1.00s) + 1/2ap(t< - 1.00s)²

                   = 0 + (26.4m/s)t + 1/2(2.5m/s²)(t - 1.00s)².

e need to consider how much time passes while the distance decreases between the two cars. Since we do not know the final speed of the police car when it overtakes the speeder, we need to consider the distance driven as it relates to time. Consider the following equivalent statements:
distance driven by speeder = distance driven by police car
distance while driving faster than police + distance while driving slower than police = distance while driving slower than speeder + distance while driving faster than speeder
distance while driving faster than police at constant speed+ distance while driving slower than police at constant speed= distance while driving slower than speeder at constant speed+distance while driving slower than speeder with acceleration+ distance while driving faster than speeder with acceleration

               xm = xp;

               (36.9m/s)t = (26.4m/s)t + 1/2(2.50m/s²)(t - 1.00s)².

          The solution to this quadratic equation are t = 0.098s and t = 10.3 s

          Because the time must be greater than 1.00s, the result is   t = 10.3s.