Car Braking Distance at Different Initial Speeds

[Mike667]

Homework Statement

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?

Homework Equations

W= ΔKE

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?

 

[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?

 

[Mike667]

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?

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?

 

[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.

 

[Mike667]

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.

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

 

[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.

 

[denverdoc]

you're getting there.
The mistake is we don't know what b is in the second case, that's an X.
So set it up as you have, and solve for X,

 

[Mike667]

Why isn't 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?

 

[denverdoc]

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

 

[Mike667]

Ummm can you show that, please?

 

[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

 

[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.

 

[denverdoc]

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

 

[Mike667]

X is 9, B is 1?

 

[denverdoc]

you have it X=9B.

 

[Mike667]

alright, thanks a lot man.

 

[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.