Solving Stopping Dist for Automobile w/ Coeff. of Kinetic Friction

[Peach]

Homework Statement

If the coefficient of kinetic friction between tires and dry pavement is mu_k, what is the shortest distance in which an automobile can be stopped by locking the brakes when traveling at v?

On wet pavement the coefficient of kinetic friction may be only mu_wet. How fast should you drive on wet pavement in order to be able to stop in the same distance as in part (a)? (Note: Locking the brakes is not the safest way to stop.)

Homework Equations

F = ma

The Attempt at a Solution

I solved the first part and I got the answer as v^2/(2*ukg). Now I'm stuck on the second part. I attempted all equations but I can't seem to find v because it cancels out on all the equations I tried.

First attempt: third eqn, vf^2 = v^2 + 2adeltax
delta x = v^2/(2*ukg)
a = 2*uwet*g
vf = 0

Also attempted to plug in first eqn of motion to find the time, and then plug in the second eqn of motion to find v but in the end, it still cancels out. Pls tell me what I did wrong...

 

[AngeloG]

First of all; I would draw a force diagram. To get a general equation with all the variables.

However, I wouldn't want a delta(x) in my final solution; that would cancel out another x when solving for distance. I would rather choose V = Vo + at. Then solve things in terms of a =).

If you still need assistance, I'll post up my solution to give you guidance but you must show me another attempt at this problem.

 

[Peach]

I drew a force diagram already, that's how I solved for the first part. For the second force, the diagram should still be the same right? Erm, but then I'd have two unknowns and one eqn...right? I solved for it in terms of a already and I have Uwetg*t. Don't I need another eqn in order to solve for t?

Edit: Unless I take that and plug it into the 3rd eqn? Nvm. I still have t.

 

[AngeloG]

Well; it doesn't matter in the end things will cancel out =). However, I would personally have chosen v = vo + at for it's convenience and ease and because it does not tell you distance. Rather just acceleration based on dV and time. It's a little redundant when you have an Xf and you're trying to solve for Xf heh on the previous step before the end.

No, you don't need another equation to solve for t. It will cancel itself out in the end to solve for the Vo for the wet pavement.

However, here is my solution: http://img102.imageshack.us/img102/5329/problemna4.jpg

 

[Peach]

OMG, I'm such an idiot. >< I got this answer awhile ago but I made the mistake of thinking that v is the same as the v I'm looking for...which it is not. I don't know what I was thinking. Thanks for taking your time to help me, appreciate it very much!

 

[AngeloG]

Haha, no problems. I have this problem; however it doesn't become clear until someone points it out to you or after hours of pulling your hair and staring at a problem. Luckily, I grow hair pretty fast =).

Example: The other day with waves and frequency stuff, I stared at a problem and tried to come up with a very complex formula to solve for something. Then after a whole night of not sleeping and the next day wasting time at the college. I realized all I had to do was divide what I was given by 4 =).

 

[denverdoc]

Why can't v'^2/2u'g=v^2/2ug where the primes denote the wet velocity and frictional coefficient;

this leads immediately to v'=v*sqrt(u'/u)? I must be missing something, here.

 

[denverdoc]

I must be missing something, I still contend that the two velocities are related to the sqrt of the frictional coefficient ratios, not directly proportional as in the soln you posted. Likey just tired, and on that note will...

 

[AngeloG]

However, I made a mistake. I believe?

In fact, time is not the same for both (I'm not sure. At first, it seem they would have to stop at the same time.) so they cannot be canceled out.

(Vo * t1 * mu_wet) / (t2 * mu_k) = Vo`

 

[PhanthomJay]

Why can't v'^2/2u'g=v^2/2ug where the primes denote the wet velocity and frictional coefficient;

this leads immediately to v'=v*sqrt(u'/u)? I must be missing something, here.

No, you are correct. As now noted, the times are not the same, only the distance is. No neeed to calculate the tme in the solution of this problem.