Using Newton's Law?

Question

A car with locked wheels stops in a distance D on a dry pavement. If the pavement is icy, so that the coefficient of how far (in terms of D) will this car travel before stopping if it started with the same initial velocity in both cases?

Answer: 8D

This equation can help solve the problem: V^2=Vo^2 - 2a(delta)x

how do I work out this problem?

Answer 1

In fact, I just worked this problem a few minutes ago for another Asker.

D or d = u^2/2kg is the stopping distance of a car going u velocity when the brakes are applied. k is the static coefficient of friction (assuming you lock the brakes) and g = 32.2 ft/sec^2 or 9.81 m/sec^2 depending on if u is fps or mps.

Using the d equation, find the ratio D/d = u^2/2Kg//u^2/2kg; where K is the dry street coefficient and k is the icy street (k < K). D and d are the distances to stop. Then D/d = k/K and D = (k/K) d is the distance on dry pavement in terms of the distance on icy pavement (d). Reverse it and d = D (K/k) the distance to stop on icy road will be greater than on dry road because K/k > 1.00

Note that u^2/2kg becomes 1/2 ut; where t = u/kg the time it takes to come to a stop while braking. And u/2, the other factor, is just the average velocity (v + u)/2 where v = 0 the final velocity when stopped.

Answer 2

The solution of the problem is a function of the energy that originally gave it motion .
If Vo^2 represents the velocity of the earth . if you throw an object in the direction of the motion of the earth it would have a velocity which depends on the aceleration and the distance it travels. In order to calculate V^2 the velocity of the earth Vo has to be set to zero as per Galileo principle of relativity.The Solution is as follows:
V^2 = 0 + 2 x acceleration x distance the mass traveled.
Where V is final velocity at that particular distance travelled.
Acceleration is the average acceleration which is half the max acceleration.
The same rules apply for a moving car.
Note if if the road is icy the only friction to stop the car is the friction of the air.
If the road is not icy and friction is present the amount of breaking energy(friction) required to stop the car would be equal to the kinetic energy that originally gave it its motion.

Answer 3

Some of the text got lost in your question, but here's the approach.

S = VoT + 1/2aT^2 or distance (S) = Initial Velocity (Vo) times Time (T) plus one half accelleration (a) times Time (T) squared. This is the standard equation for calculating distance traveled.

F = ma or Force equals Mass (m) times accelleration (a) pure Newton.

So, given how far the car took to stop on dry pavement, (D) we solve equation 1 for the accelleration which is also equal to F/m by equation 2. The Force is the coefficient of friction times the normal force which is mg where g is the accelleration of gravity, so we can substitute and solve the two equations in terms of D.

Answer 4

Start by considering the work done by friction to stop the car:
f*D, where f is the frictional force and D is the distance to stop. This force will convert all of the kinetic energy of the car to heat, which means we can write the equation
.5m*v0^2=f*D

Now consider the equation for f:
The normal force, N (m*g), times the coefficient of friction, u:
f=N*u
so
D=.5*m*v0^2/(N*u)
you stated that the initial velocity, v0 will be the same, so let's set up a constant, k, where
k=.5*m*v0^2/N

so now we can express D as a function of u:
D(u)=k/u

your post doesn't say what the change in u is, but based on the answer, the u for icy versus dry is 1/8 or
u icy=u dry/8

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