# Velocity problem # 2

1. Feb 5, 2009

### lauriecherie

1. The problem statement, all variables and given/known data

On a dry road, a car with good tires may be able to brake with a constant deceleration of 4.06 m/s2.

(a) How long does such a car, initially traveling at 23.6 m/s, take to stop?
5.813 s This answer is correct.

(b) How far does it travel in this time?
I've tried 59.47 m and 137.1868 but Webassign says both are incorrect. I have no clue as to what I am doing wrong.

2. Relevant equations

x(t)= initial position + final velocity * time

v(t)= (acceleration * time) + initial velocity

x(t)= .5 * (acceleration * (time^2)) + (initial velocity * time) + inital position

x= initial position * (average velocity * time)

average velocity= (final velocity - initial velocity) / (2)

(final velocity^2) - (initial velocity^2) = 2 * acceleration * change in position

3. The attempt at a solution

I got 59.47 meters by pluging in part a, 5.813 seconds, into the position function.
Then I tried 137.19 meters after figuring that in 5.813 seconds at 23.6 m/s (I multiplied the two) it could travel 137.19.

Any ideas on what I am doing wrong?

2. Feb 5, 2009

### Delphi51

x(t)= .5 * (acceleration * (time^2)) + (initial velocity * time)
= .5*(-4.06)*...............
The acceleration is negative because it is slowing the car. Decelerating.

3. Feb 5, 2009

### lauriecherie

68.59 meters. Thanks! I really REALLY appreciate it. I'm seeing the little things I'm doing wrong now.

4. Feb 5, 2009

Super!