# Physics - proving the stopping distance of a car

1. Apr 24, 2016

### totomyl

1. The problem statement, all variables and given/known data
an accident occurs up ahead on the highway. a driver travelling at 120km/h [e] reacts in 0.20s and applies the brakes causing an acceleration of 8.0m/s2 [w]. show that the stopping distance is 76 m.
what am i doing wrong? i changed the acceleration to match the directions, so i made it negative. but i am not getting the right answer.

2. Relevant equations
d = vi * t + 0.5(a * t^2)

3. The attempt at a solution
i attempted this by using:

d = (120km/h / 3.6[e])(0.20s) + 0.5(-8.0m/s[e] * 0.20s^2)

d = 6.5m?

2. Apr 24, 2016

### SteamKing

Staff Emeritus
The problem is you have assumed that the car goes from 120 kph to 0 kph in 0.2 s, which is not what the problem states. The driver takes 0.2 s to press the brake pedal after he sees the accident ahead of him.

You should pick another SUVAT equation which relates distance, acceleration,and initial and final velocity.

3. Apr 24, 2016

### totomyl

So, I just have another question, in this question the final velocity would be at a stop, so 0 m/s correct? and also would i have to find the distance traveled before pressing on the brakes and add it to the distance it took while slowing down?

4. Apr 24, 2016

### SteamKing

Staff Emeritus
Yes and yes.

5. Apr 24, 2016