Finding the displacement of a braking car, with limited variables.

[Symon]

Homework Statement

A car is traveling at 120km/h when it slams on the brakes. How long is the skid mark if the coefficient of friction is 0.62?(hint: convert km/h to m/s)

Given: initial velocity: 33.3m/s; final velocity: 0m/s; coefficient of fricition: 0.62

Homework Equations

μ = Ffriction/Fnormal; a= Force/mass

The Attempt at a Solution

Because of the absense of a time value, I'm finding it very difficult to use and displacement equation I've been given. I understand the the change in velocity (after being conversted to m/s) is 33.33m/s, but that's as far as I've gotten.

I was wondering is anyone could help me derive some other variable: such as the change in time, or the acceleration, with the given data.

 

[tiny-tim]

Welcome to PF!

A car is traveling at 120km/h when it slams on the brakes. How long is the skid mark if the coefficient of friction is 0.62?(hint: convert km/h to m/s)
…
Because of the absense of a time value, I'm finding it very difficult to use and displacement equation I've been given.

Hi Symon! Welcome to PF!

Is the "displacement equation" work done = force times (or "dot") displacement?

If so, you don't need the time.

 

[Symon]

Thus far, i am unfamiliar with that equation, here are some of the equations that I've either derived or have been given:

1) ∆d=〖(∆v)〗^2/2a
2) ∆d= ∆v∆t
3) ∆d=vi∆t+ 1/2 a〖(∆t)〗^2

As you can see, the first equation that i have neglects time, but i am unable to use it without finding the acceleration (which is another equation i need time for). Is there some alternate way i can find the acceleration, force involved, time, or mass?

 

[tiny-tim]

ok … you're using the usual constant acceleration equations.

(but #1 is wrong … it should have ∆(v²), not (∆v)², and #2 is completely wrong … replace it with ∆v = a∆t)

Yes, you can find the acceleration, a, by starting with the friction coefficient, µ = 0.62 …

that will give you the force, and then you can find the acceleration from good ol' Newton's second law.

 

[Symon]

Thank you :)