# Old exam question

1. Apr 16, 2012

### Mdhiggenz

1. The problem statement, all variables and given/known data
A police car at rest is passed by a speeding car moving east at 20m/s. The police officer immediately starts to accelerate to the east at 5.0m/s. How long will it take the police officer to catch up with the offending car

2. Relevant equations

3. The attempt at a solution

VIcop=0
VFcop=?
Acop=5.0m/s2

VIoffender=20m/s

I dont really know which equation to start with because I feel as if I dont have enough variables. I want to solve for the final velocity of the cop then use that to get the time by making another equation where the final of the cop, becomes the initial, and the initial of the offender is the final.... ahh im so confused.

2. Apr 16, 2012

### Staff: Mentor

Don't focus on the velocities, focus on the positions. The positions of the two vehicles will be the same twice: once at time zero when the driver passes the stationary cop car, and again when the cop car has caught up with driver. The speeds are not the same at either time.

3. Apr 16, 2012

### november1992

The way I did it was by setting the equations equal to each other

So, since the car is moving 20m/s and the cop car is accelerating 5m/s^2:

vt=1/2at^2

Last edited: Apr 16, 2012