Assuming the speeder keeps a constant speed of 100 km/h, equaling 27.78 m/s, his position from origo can be expressed as an integral with the solution
d1 = v* t
d1 = 27.78* t
Also assuming that the police keeps a constant velocity of 3.6 m/s2 (no matter the speed), and having a starting velocity v0=0 m/s then the position of the police car from origo can be found by solving the corresponding double integral. The solutioin of the integral is.
d2 = v0*t + 0.5a*t2
d2 = 0 + 0.5*3.6*t2 = 1.8 t2
The moment when the cop passes the speeder will be given as the intersection point between the two distance functions.
d1=d2
1.8 t2 = 27.78 t
Which can easily be solved for t.
1.8 t = 27.78 --> t = 27.78 / 1.8 = 15.433
During those 15.5 seconds, both vehicles have travelled a distance equal to
d1 = 27.78 m/s * 15.433 s = 428.7 m
and
d2 = 1.8 m/s2 * (15.9433 s)2 = 428.7 m
Let's check the final speed of the police car. The speed will be given as
v2 = 3.6 * 15.433 = 55.5588 m/s = 200 km/h
This is exactly double of the speeding car's velocity. That makes very much sense. Assuming the police car has constant acceleration, that means that the velocity of the police car grows linearly (basic integration). The linear velocity function of the police car goes through 0 at the start, so in order to have the same area under the curve (integral = distance travelled) as the car with constant velocity, the final velocity of the police car must be twice the velocity of the speeding car. This can be easily visualized by graphing the two velocity functions in the same plot.
