# Constant Acceleration

1. Jan 31, 2006

### blanny

Hi, this one is probably easy, but I just can't wrap my head around it.

You are arguing over a cell phone while trailing an unmarked police car by 25m; both your car and the police car are traveling 110km/h (about 31m/s). You take your eye off the road for 2.0s because of the argument. At the very beginning of the 2.0s, the police car brakes at 5m/s^2

a) what is the separation between the two cars when your attention returns?

I worked it out and I got about 20m. Is this about right?

b) Suppose you take another .40s to realize the danger and begin braking. If you too brake at 5m/s^2, what is your speed when you hit the police car?

I'm not too sure how to set this one up. I'm thinking it has to be a system of equations...

after 2.4s the distance should be 19m, but how do I set up the system from there?

The equations in the chapter are the equations for motion with constant acceleration.

thanks

Matt

2. Feb 4, 2006

### Staff: Mentor

No. How did you arrive at that answer? [Note: If you show your work, you'll get better help quicker!]

One way to solve it is to set up equations for the position as a function of time for each vehicle, then see where they are after 2 seconds.

That distance is wrong; solve for it using the same equations as used for part a (which you need to redo).

To solve part b, first find the position and speed of both vehicles at the moment you begin braking. Then set up new equations for each vehicle and solve for the moment that they collide. (Once you have the time, you can figure out the speed.)