Calculating Car Speed: Using Skid Marks and Friction Coefficient

[TriumphDog1]

ok, i have a problem and its answer, but I am not sure how to get the answer:
Police lieutenants, examining the scene of an accident involving two cars, measure the skid marks of one of the cars, which nearly came to a stop before colliding, to be 81 m long. The coefficient of kinetic friction between rubber and the pavement is about 0.70. Estimate the initial speed of that car assuming a level road.
the answer is 33.34.
I can't figure out how to get that answer.

 

[Evgeny]

Essentially, this is a problem with requires the use of kinematics and Newton's second law.
The only force acting on the car (neglecting things like drag) is the force of the friction, F_k
Now, we know that F_k = \mu _k N
Here, the normal force N = mg
Now, let us substitute N into F_k:
F_k = \mu _k mg
From Newton's second law we know that: F_{net} = ma
But since we know that the only force acting on the car is F_k, we can substitute F_k for F_{net}. We get:
F_k = ma
\mu _k mg = ma
Canceling the m-s out,
\mu _k g = a
Now, let us use kinematics equations:
v_f^2 = v_0^2 + 2a(\Delta x)
From which we get,
v_0 = \sqrt{2a(\Delta x) - v_f^2}
Substituting for a
v_0 = \sqrt{2(mu _k g)(\Delta x) - v_f^2}
Substituting our givens,
v_0 = \sqrt{2(0.70)(9.81)(81) - 0}
v_0 = 33.34
QDE

Hope this helped.

 

[TriumphDog1]

yes thank you.