   SEARCH HOME Math Central Quandaries & Queries  Question from Joel, a student: A car is going down the road at a speed of 90km/h in a 50km/h zone. As the car passes a corner, a police cruiser sitting there notices it. Five seconds later the cruiser begins accelerating at 5.0m/s squared. Assuming that he is able to maintain this acceleration, how far from the corner will the police catch the car? How fast will the cruiser be going? Hi Joel.

You want to know the distance from the corner, so calculate for distance for each vehicle:

Car: d = vt + (1/2)at2 = (90 km/hr) t + (1/2)(0) t2

Cruiser: d = vt + (1/2)at2 = (0) t + (1/2)(5 m/s^2) t2

You are concerned with the point in time that they are together (the same distance from the corner). So the d values are the same. As well, we started both clocks when they were together at the corner, so the t's are the same.

Therefore, we can make one equation equal the other:

Car distance = Cruiser distance
(90 km/hr) t = (1/2)(5 m/s^2) t2

Solve for t and you get:

2 x (90 km/hr) / (5 m/s2) = t

The rest is unit conversion. Once you know the time, you can use this value for t in either of the original equations to find the distance.

Then you can use the equation v = at to find the speed of the cruiser when the police officer overtakes the speeder.

Cheers,
Stephen La Rocque.     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.