Complicated Physics problem will choose best answer?

Question

A car is traveling at 50.0 km/h on a flat highway. If the coefficient of friction between road and tires on a rainy day is 0.100, what is the minimum stopping distance in which the car will stop? What is the stopping distance on a dry day when the coefficient is 0.600?

Answer 1

Actually this is not that complicated.

The frictional force on the car is F = μN = μ(mg), since the Nornal force on the car is equal to mg and μ is the coefficient of friction

Now use F = ma = μ(mg) implies that the car's acceleration is:

a = μg

Now μ = 0.1 during a raining day and μ = 0.6 on a dry day

to find the stopping distance, we use the standard equations of motion:

v(f)² = v(i)² - 2ad

v(f) = 0 per requirement that the car comes to a stop, then

v(i)² = 2ad = 2μgd, solve for d

d = v(i)²/2μg

For μ = 0.1, d =(13.89)²/2(0.1)(9.8) = 98.42 meters

For μ = 0.6, d =(13.89)²/2(0.6)(9.8) = 16.41 meters

Answer 2

You don't know from 'complicated' ☺

If the coefficient of friction is .1, then the maximum decelleration is .1g or .98 m/s². If it's .6 then the decelleration is .6g = 5.88 m/s². If you look in your physics book (damn! what a concept), you'll find an equation that says something like
V² = v² - 2ax where V is final velocity, v is initial velocity, a is decelleration, and x is distance (over which the decelleration is applied) To stop means that V is zero, so the equation becomes 0 = v² - 2ax or (with a bit of algebra) x = v²/2a.

Remember to convert 50 km/hr to 13.889 m/s.

On a wet day x = 13.889²/(2*.98) = 98.32 meters. You work out the dry day ☺


Doug

Answer 3

hi those questions are previous information these days with ABS braking structures as they in lots of circumstances go away no skid mark any further what you ought to work out in spite of the incontrovertible fact that are marks suggesting a collision by way of bits of the automobile alongside area of highway from the place the collision happen ed.

Answer 4

Min stopping distance on a raining day(s)=(964.5/g)m i.e
if g=10m/s^2 then s=96.45m
else if g=9.8m/s^2 then s=98.41m

Min stopping distance on a dry day(x)=(160.75/g)m i.e
if g=10m/s^2 then s=16.07m
else if g=9.8m/s^2 then s=16.40m

Answer 5

98.32 m (322 ft) in the 1st case (mu = 0.1)
and 16.39 m (53.816 ft) in the 2nd case (mu=0.6)

Answer 6

37.6676436595836539265m