# Stopping distance

1. Aug 24, 2010

### Travian

1. The problem statement, all variables and given/known data
The maximum straight-line deceleration of a racing car under braking is 5 m s-2. What is the minimum stopping distance of the car from a velocity of 108 km h-1?

2. Relevant equations

½ V x t

3. The attempt at a solution

V = 108km/h=30m/s

t = 30/ 5 = 6 m s-2

½ V x t = ½ x 30 x 6 = 90

___________

I found this formula ½ V x t (to determine stopping distance) on the internet, but im not sure if its correct.

Are there any mistakes?

Thank You

2. Aug 24, 2010

### Herr Malus

It seems like your formula for distance is the average distance travelled in a given time for an object moving at constant speed. I would Instead try the formula:
vf 2 = vo 2 + 2 a $$\Delta$$ x

Note that you are decelerating so the plus in front of the 2 is actually a minus here.

Last edited: Aug 24, 2010
3. Aug 24, 2010

### Travian

so: vf = 30 m/s -2 x 5 m/s...
what is this delta x? sorry, haven't slept this night at all, feeling a bit dizzy now:/

4. Aug 24, 2010

### Herr Malus

Sorry, delta x here stands for the change in position of the car.

Basically, how far it goes in stopping.

5. Aug 24, 2010

### Travian

90 meters?

6. Aug 24, 2010

### Herr Malus

Yes, 90 meters. Aside from the units in your 't', everything seems to be in order.

7. Aug 24, 2010

### Travian

Thank you.

By the way, whats wrong with t?

8. Aug 24, 2010

### Herr Malus

Units: I'm a little mixed up by your scheme but it looks like you have t (time) in meters per second per second, instead of just seconds.

9. Aug 24, 2010

### Travian

Oh seriously... I fixed that. Thanks again:)