# Car braking distance question

1. Apr 24, 2007

### Mike667

1. The problem statement, all variables and given/known data
A car with an initial speed of 20 m/s brakes to a complete stop after traveling distance b from the moment the brakes are applied. What would be the minimum braking distance if the initial speed was 60 m/s?

2. Relevant equations

W= ΔKE

3. The attempt at a solution
I know for a fact that the answer is 9 or 9b. Simply because 60 is 3 times greater than 20. 3^2 is 9. But, I am having trouble solving this with the formula. Can someone help me out here?

2. Apr 24, 2007

### denverdoc

Two basic ways to approach this; from kinematics there is an eqn which relates velocity^2 to a, acceleration and x distance.

An alternative approach would be to use work energy theorum, which is how you are setting this up. Do you know an eqn that relates frictional force (braking in this instance) with distance?

3. Apr 24, 2007

### Mike667

ΔKE is 1/2(m)(v)^2 right? I have no idea what you mean by an equation that relates frictional force with distance. f*d, perhaps? Where would i put the b in the formula?

Last edited: Apr 24, 2007
4. Apr 24, 2007

### denverdoc

you're on the right track, my man. B would go where d is.

since final velocity is zero, we know all energy went into friction unless there was a vertical difference, say up a slope involved.

so we set up up eqn, relating the initial kinetic energy to a stopping force, which is directly proportional to mass as well:

1/2mV^2=Ff*b where Ff=m*g*mu, my lord you say we now have more variables. The way out and probably the easiest approach is to set up two equations representative of each situation and to divide the two.
Hope this helps.

5. Apr 24, 2007

### Mike667

g is 10 right? but what does mu stand for?

6. Apr 24, 2007

### Mike667

Ok, so

f*b = 1/2(m)(20)^2

f*b = 200m?

f*b= 1/2(m)(60)^2

f*b= 1800m?

I feel like I am missing something here.

7. Apr 24, 2007

### denverdoc

you're getting there.
The mistake is we dont know what b is in the second case, thats an X.
So set it up as you have, and solve for X,

8. Apr 24, 2007

### Mike667

Why isnt b an x in the first case then?

f*b = 1/2(m)(20)^2

f*b = 200m?

f*x= 1/2(m)(60)^2

f*x= 1800m?

Then what? how can i get the answer to just be 9b?

9. Apr 24, 2007

### denverdoc

perfect, divide the two eqns, and solve for x in terms of b.

10. Apr 24, 2007

### Mike667

Ummm can you show that, please?

11. Apr 25, 2007

### denverdoc

I think I see a problem here, you have problems with word problems.
Not alone! Your algebra could likely benefit as well from some tutoring.

So on top of the fraction we have:
f*b=200 on bottom we have fx=1800
therefore b/x=1/9

12. Apr 25, 2007

### Mike667

Wait wait wait, so you're saying the answer is 1/9, not 9b?
I definitely see where you're coming from, its just that I supposed the answer would look different.

13. Apr 25, 2007

### denverdoc

B/x=1/9, then what is X?

14. Apr 25, 2007

### Mike667

X is 9, B is 1?

15. Apr 25, 2007

### denverdoc

you have it X=9B.

16. Apr 25, 2007

### Mike667

alright, thanks a lot man.

17. Apr 25, 2007

### denverdoc

No sweat, recommend PF to all! And get some assistance with algebra if you can as this was the biggest barrier to the soln. Your reasoning spot on, just unfamiliarity with tricks to get the answer.