# Stopping Distance

1. Oct 10, 2004

### quick02si

Hello everyone this is my first time in the forum and i'm actually having lot of trouble with my Physics homework. Hope someone can help me. Here's one of the problems and thanks in advanced.
If the coefficient of kinetic friction between tires and dry pavement is 0.800, what is the shortest distance in which an automobile can be stopped by locking the brakes when traveling at 29.7 ? Take the free fall acceleration to be 9.80 .

Last edited: Oct 10, 2004
2. Oct 10, 2004

### Sirus

Please include units with all your data whenever you post. Also show us how you have attempted the problem and where you got stuck.

3. Oct 10, 2004

### quick02si

I'm sorry I didn't even notice that I missed the units.
Here is the problem again:
If the coefficient of kinetic friction between tires and dry pavement is 0.800, what is the shortest distance in which an automobile can be stopped by locking the brakes when traveling at 29.7m/s? Take the free fall acceleration to be 9.80m/s^2.
On this problem I don't know how to start it off. Thats where I get stuck. I don't know how to incorporate the kinetic friction to find distance.

4. Oct 10, 2004

### Sirus

I think this is best done with work-energy concepts due to the lack of a value for mass of the car. What happens to the initial (kinetic) energy of the car while it skids?

5. Oct 10, 2004

### quick02si

I think I see what you are saying but my problem is that I haven't completely gone through that yet. That is the next chapter. The chapter that is problem comes from is Applying Neton's Laws, but i'm stuck. Thanks for your help

6. Oct 10, 2004

### Sirus

So they want you to do this with kinematics formulas?

7. Oct 10, 2004

### quick02si

Yes, thats really the only thing i know. But still can't get the problem.

8. Oct 10, 2004

### Sirus

You can do it if you know newton's law $F_{net}=ma$. What is the net force acting on the car during the skid in the plane of motion (ignore normal force for now)?

9. Oct 10, 2004

### quick02si

ok so the sum of the forces in the x direction would be the velocity-coefficient of kinetic friction=ma

10. Oct 10, 2004

### Sirus

I don't think so. The only force acting on the car in the plane of its motion is the frictional force. Since $F_{kinetic friction}=\mu_{k}F_{N}$, and $F_{N}=mg$, you should be able to use F=ma and this information to find the acceleration of the car. Can you take it from there?

11. Oct 10, 2004

### quick02si

ok so mu_{k}(mg)=ma so a=mu_{k}(g). Is this what you are trying to get me to see.

12. Oct 10, 2004

### Sirus

Correct. Now consider the information you have, and apply an appropriate kinematics formula to find the distance.