Calculating Initial Velocity in a Simple Car Braking Problem

[Xeonicus]

The problem is simple, its a matter of finding a car's initial velocity after locking up the wheels while braking. Here is what I know:

The braking distance (skid marks) go for 20 feet, the final velocity is 50m/s, the mass of the vehicle is 2000kg, and the coefficient of friction of the tire and asphalt is 0.6. What is the initial velocity?

Here is how I did it first:
1) force friction = 0.6 * 2000 * 9.8 = 11760N
2) m * a = 2000 * a = -force friction + ( 2000 * -8.5 )
2000 * a = -11760 - 17000
a = 14.38 m/s^2
3) Then I go on to solve for the intial velocity after getting the deceleration.

In step 2, I used -8.5 for a typical deceleration rate of a vehicle when it slams on the brakes. What I'm told though, is that I don't need to add this force for braking and I only need to put in the friction force (0.6 * normal). So I'd just have "m * a = -force friction" and the deceleration would be -5.88 m/s^2. It seems to me though, that a vehicle would deceleration more quickly if the brakes were applied, more so than just slowed down by the force of rolling friction the road applies.

So can anyone clear this up for me? Something to do with internal/external forces?

 

[Michael D. Sewell]

1.) Lock up wheels with brakes = coefficient of sliding friction

2.) Brake as hard as possible without skidding = coefficient of static friction

In an emergency-if you like your car, use #2.

 

[HallsofIvy]

"In step 2, I used -8.5 for a typical deceleration rate of a vehicle when it slams on the brakes. What I'm told though, is that I don't need to add this force for braking and I only need to put in the friction force (0.6 * normal). "

I was wondering where that "8.5" came from! The only force causing the car to decelerate IS the friction of the tires sliding on the road.

As long as the wheels are rolling, the road does not supply ANY "rolling friction".

 

[jdavel]

HallsofIvy said: "As long as the wheels are rolling, the road does not supply ANY "rolling friction"."

If it doesn't when the wheels are rolling, then when does it?

Don't you really mean that since rolling friction is much less than static friction, that during breaking, if the wheels are rolling, the frictional force applied by the road to the car is nearly equal to the static friction?

 

[Michael D. Sewell]

HallsofIvy said: "As long as the wheels are rolling, the road does not supply ANY "rolling friction"."

If it doesn't when the wheels are rolling, then when does it?

Don't you really mean that since rolling friction is much less than static friction, that during breaking, if the wheels are rolling, the frictional force applied by the road to the car is nearly equal to the static friction?

Sorry, this is my fault, my original post said "rolling friction". I should not have
mentioned rolling friction at all. The rolling friction is not a factor in this problem, only the static friction or the sliding friction. I edited it so as not to cause confusion for anyone else. I must have been on crack when I wrote that! -Mike