Stopping distance of a car

1. Oct 27, 2007

bigtymer8700

The driver of an 1800kg car traveling at 29.0m/s slams on the brakes, locking the wheels on the dry pavement. The coefficient of kinetic friction between rubber and dry concrete is typically 0.600. Find the stopping distance

ive been trying to this problem forever and i know that Its just im missing something small but i cant figure out what.

Ive solved for the KE_initial= 756900J and i know K_final= 0 I know i have to use the constant accel. equation
vf^2=vi^2 + 2as. I just dont know how to go about starting the problem all the unknowns have got me confused.

My attempt went like this. I got the F_normal by 1800kg(9.8) to get 17640N.
17640(.600) gave me 10584N which i used as the frictional force.
a=F/m so then 17640N-10584N/1800 =3.92 i used that for acceleration but i got the wrong answer i dont know where im messing up

Last edited: Oct 27, 2007
2. Oct 27, 2007

Poop-Loops

Energy isn't important in this case. Only speed, mass, force, distance and acceleration. You should have a formula relating them.

You have initial and final speed and mass already. You can find the frictional force. It's the coefficient of friction times the normal force. You know what the normal force is, right? It's the weight of the car.

Now, the frictional force is what's stopping the car, right? So if you know it and know the mass of the car, you can find the deceleration, and you should have a formula to get distance from all of that.

3. Oct 27, 2007

bigtymer8700

dont you need the acceleration to get the normal Force?

Last edited: Oct 27, 2007
4. Oct 27, 2007

Poop-Loops

Nope. The mass of the car is 1800kg. To find the force you have to multiply by acceleration. Think of which way the normal force is pointing.

5. Oct 27, 2007

bigtymer8700

Normal force is pointing up i got that from my free-body diagram so i did 1800kg(9.8) to get 17640N

6. Oct 27, 2007

Poop-Loops

You're off by a zero, but yeah, that's it.

So you have the normal force. So F = u*m*a. That's your stopping force.

7. Oct 27, 2007

bigtymer8700

yea I got that far 17640(.600)= 10584N for the stopping force. but how do i go about solving for the acceleration to get the stopping distance?

Last edited: Oct 27, 2007
8. Oct 27, 2007

bigtymer8700

Ok I got some insight on the problem and hopefully this is the way. but i know that W= K_f - K_i. I already know the car is stopping so K_final is 0.
K_i is (.5)1800kg(29m/s)^2 which gives you 756900J. Since you know Work is also W=F * s you got the Work from the KE and F=17640N. you divide 756900J/17640N to get 42.9 is that correct?

9. Oct 27, 2007

Poop-Loops

Yes, that works, but I don't understand why you insisted on using energy here. It works here because it's a simple system. If you get something more complicated, energy probably won't be "conserved", so I'd be careful and stick to forces instead.

10. Oct 27, 2007

hage567

Shouldn't you be dividing by 10584 N here? That is what the frictional force was found to be. 17640 N is the normal force.

11. Oct 27, 2007

bigtymer8700

so to get that distance you have to divide 756900/10584?

12. Oct 27, 2007

hage567

Yes. Do you understand why? The frictional force is what is causing the car to stop, not the normal force.