# Kinematics car chase question

1. Oct 18, 2007

### spizma

1. The problem statement, all variables and given/known data
A speeder passes with velocity 100 km/h by a police car with velocity 80km/h. Exactly one second after the speeder passes the police car, the police car accelerates with an acceleration of 2 m/s^2. How long is it until the police car passes the speeder.

2. Relevant equations
$$x = x_0 + v_0 + \frac{1}{2} a t^2$$

100 km/h = 27.78 m/s
80 km/h = 22.23 m/s

3. The attempt at a solution
I figured out that in the one second before the police car accelerated, the speeder got 5.55 m ahead. I then figured that the distance traveled by the police car from the time it starts accelerating to the time it catches up with the speeder (t) is 5.58 m + (27.78 m/s)*t (the distance between the cars at first, and the distance traveled by the speeder in the time it takes for the police to catch up).

So the equation should be this, right?
$$5.55 + 27.78 t = 22.23 t + \frac{1}{2} 2 t^2$$

When I try to solve for that I get t = -.87 s and 6.42 s, but the back of the book says it is 7.4 seconds. What am I doing wrong? Thanks much in advance.

Last edited: Oct 18, 2007
2. Oct 18, 2007

### Vijay Bhatnagar

Your calculation is correct. 6.42 sec is the right answer. 7.4 sec will be the solution if t=0 is the instant when speeder passes by the police car.