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Question

At an accident scene on a level road, investigators measure a car's skid mark to be 88m long. THe accident occurred on a rainy day, and the coefficient of kinetic friction was estimated to be 0.42. Use these data to determine the speed of the car when the driver slammed on (and locked) the breaks. (Why does the car's mass not matter?)

Answer 1

The force of friction F equals:

F = c*m*g

where c: the coefficient of friction, m: the mass of the car and g: the acceleration of gravity (= 9.81 m/s^2 approx.)

Now you can calculate the deceleration of the car, according to the Newton's second law:

F = m*a => a = F/m, or a = (c*m*g)/m. So:

a = c*g (That's why the car's mass not matter).

The maximum distance Xmax that the car travels before stop is given by the equation:

Xmax = v^2/2*a

where v the velocity of the car before deceleration, so:

v = sqrt(2*Xmax*a) or v = sqrt(2*Xmax*c*g)

v = sqrt(2*88*0.42*9.81), v = sqrt(725.16) and, at last:

v = 26.93 m/s (approx.)

Answer 2

When a car is decelerated the change in its velocity is due to the force of friction T = n * A, where n is the friction coefficient and A the vertical force on the common surface of the two objects (in this case the road and the car).
Since the road is level then A = m*g i.e. the weight of the car.
So T = n * m * g
Applying Newton's second law (Ftotal=m*a) we get
T = m * a -> n * m * g = m * a
As we can see we can cancel mass m, so that
a = n * g (1)
We also know that
v = v0 - a*t (2)
and
s = v0*t - (1/2)*a*t^2 (3)
where v is the velocity at time t and v0 the initial velocity.
When the car stops then obviously v=0
Solving (2) for time t we get

t=v0/a (4)

Combining (3) and (4) we have

s = v0*(v0/a) - (1/2)*a*(v0/a)^2 ->
s = v0^2/a - (1/2)*v0^2/a ->
s = v0^2/(2*a) (5)

Combining (1) and (5) we get
s = v0^2/(2*n*g) (5)
Solving for v0 we get
v0 = SQRT[ 2*n*g*s ]
where SQRT stands for square root
Assuming g=10m/s^2 we have the result
v0 = 27.19 m/s
(approximate value)

Answer 3

about 12

Answer 4

Let the mass of the car be m kg
Its weight=mg Newtons. where g is accn.due to gravity
Frictional force acting on the car=.42mg N
Deceleration( linear)=force/mass=.42m*9.81/m
=.42*9.81
(we observe m cancels out.So it does not matter)
v^2-u^2=2*-a*s
v=o u=? s=88m a=.42*9.81
u^2=2*.42*9.81*88
u=26.96m/sec=97kmph

Answer 5

The car's mass cancels out of the calculations because the frictional force and the momentum are inversely dependent on it.

Answer 6

I'm not sure of the answer, but I would think the cars mass would have to matter.

Answer 7

above 115 km/h