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.