Physics Help !?

Question

A car is traveling at 33.4 mi/h on a horizontal highway. The acceleration of gravity is 9.8 m/s2. If the coefficient of friction between road and tires on a rainy day is 0.1 what is the minimum distance in which the car will stop ? (1mile = 1.609 kilometers )

Part B) What is the stopping distance when the surface is dry. Coefficient of when its dry is 0.669

Answer 1

( NOTE: As you can see below in the source link, Google Calculator does not agree with your conversion that 1 mile = 1.609 kilometers. However, I have used your conversion factor to be consistent with your problem. If you use Google's conversion factor, you will get a slightly different answer )

Assume the mass of the car is m and the acceleration due to gravity is g (so g=9.8 m/s/s). The weight of the car is thus m*g. That means the magnitude of the normal force pushing up ont he car is also m*g. So the frictional force pushing back on the car is mu*m*g (where mu in the first part is 0.1). If we want to find the acceleration due to the friction, we divide by the mass m. That means the acceleration due to friction is sipmly mu*g.

You know the initial velocity of the car going into the stop is (33.4 mi/hour)*(1.609 kilometers/mile)*(1000 m/kilometer)*(1 hour/60 minutes)*(1 minute/60 seconds) = 14.9279 m/s.

You also can set the initial position to be 0 meters.

Now, you want to solve for the time it takes to stop the car. Since acceleration is just the rate of change of velocity, then you know the velocity of the car is given by:

v = a*t + v0

where a = -mu*g and v0 = 14.9279 m/s. You are solving for the time t when (v = 0 m/s). That is, you are solving the following equation for t:

(0 m/s) = -mu*g*t + 14.9279 m/s

which gives:

t = (14.9279 m/s)/(mu*g)

Now, all you have to do is figure out the distance traveled in this amout of time. The distance formula is:

x = 0.5*a*t^2 + v0*t + x0

Where a = -mu*g, v0 = 14.9279 m/s, and x0 = 0 meters. That is:

x = (14.9279 m/s)*(14.9279 m/s)/(mu*g) - 0.5*mu*g*( (14.9279 m/s)/(mu*g) )^2

You can write that a little more simply as:

x = 0.5*(14.9279 m/s)^2/(mu*g)

Now you can solve for both of your answers. You just have to plug in different values of mu, the coefficient of friction.

Part A) (with mu=0.1) x = 113.696 meters

Part B) (with mu=.669) x = 16.9949 meters

As you can see, the car stops much more quickly when it slides on dry land (with a higher coefficient of friction).

Answer 2

in real life.. u need the radius of the car tires.. but if you consider the car to be a block then

33.4mi/h = 14.93 m/s

F = m(dv/dt) = mv(dv/dx)
mv(dv/dx) = -μN
mvdv = -μNdx

integrating the stuff we get

(1/2)mv^2 = μNx
x = (1/2μN)mv^2
x = [1/(2*0.1*N)]*m*(14.93^2)

but N = mg so

x = [1/2*0.1*g)]*(14.93^2)
x = [1/2*0.1*9.81)]*(14.93^2)
x = 113.58m

when it is dry, just change 0.1 to 0.669 so

x = 16.9775m

Answer 3

要获得最小的制动距离．得经过三个过程，匀加速，匀速，匀减速，首末过程的加速度为0.98 m/s2。中间过程实际上没有经历．then left is the pen-work.
the minimum distance is s =2×(1/2 v^2/a) =v^2/ug===? .you can make it .