One more....for now?

Question

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 77 m long. The coefficient of kinetic friction between rubber and the pavement is about 0.80. Estimate the initial speed of that car assuming a level road.

Answer 1

The force of friction between the car’s tires and the road is doing work on the car to change/lower its Kinetic Energy.
Since the car nearly came to a stop as it skid 77 meters down the road, the final KE value was about zero, so the total change in KE is about the initial value before the breaks were applied and the car (of mass m, traveling at speed v),
KE = ½ mv^2

The force of friction did work on the car over a distance,
Work = Force * Distance
Force of friction = coefficient of friction * mass * gravity

Setting the work done by friction equal to the change in KE,

½ * m * v^2 = μ * m * g * d
Where m is the mass of the car, v is the car’s initial speed, μ is the coefficient of friction between the tires and the road, g is the gravitational acceleration experiences by the car, and d is the distance over which the car stopped.

Since the mass of the car appears on both sides of the equation, we can cancel it out without affecting the result.
We do not actually need to know the mass of the car in order to solve this problem since it cancels out.

We can solve for v to be,
v^2 = 2 * μ * g * d
v = sqrt (2 * μ * g * d)
Plugging in the values that we know,
v = sqrt (2 * (.8) * (9.81 m/s^2) * (77 m))
v = 34.8 meters per second

The car’s initial speed before slamming on the breaks was about 34.8 meters per second, or about 77.8 miles per hour.

Answer 2

As the required time to stop just about any vehicule is about two seconds, it will travel two seconds worth of it's initial velocity.

If it took 77 m to stop then it had to be going roughly 38 meters per second.

Answer 3

Using

V^2 = Vo^2 + 2ax

V: final velocity ("0" car has come to a stop)
Vo: initial velocity (TO SOLVE FOR)
a: acceleration, (deceleration here)
x: distance (of acceleration/deceleration, the SKID MARKS)

ALSO:
From F=ma we have: a = F/m
the F (force) causing the car to come to a stop is the that due to friction between the road and the tires,.. frictional force Ff.

Ff = uf(With of the car) = uf(Mass of Car)(Acc. Gravity)

Now:
a = uf(Mass of Car)(Acc. Gravity)/(Mass of Car)

After simplifying and inserting values:

a = (0.8)(9.8 m/s^2)

Then this into: V^2 = Vo^2 + 2ax, we go (-2ax since DECELERATING)

0 = Vo^2 - 2(0.8)(9.8 m/s^2)(77 m)

Rearranging:

Vo^2 = 1207.4 (m/s)^2
Vo = 34.7 m/s (approx. 125 km/hr.)

Answer 4

92 mph