To test the performance of its tires, a car

travels along a perfectly flat (no banking) cir-

cular track of radius 516 m. the car increases

its speed at uniform rate of

at ≡d |v|÷dt = 3.89 m/s^2

until the tires start to skid.

if the tires start to skid when the car reaches

a speed of 32.8 m/s, what is the coefficient of

static friction between the tires and the road? the acceleration of gravity is 9.8 m/s^2

Answer:

The coefficient of  static friction between the tires and the road is 1.987

Explanation:

Given:

Radius of the track, r =  516 m

Tangential Acceleration =  3.89 m/s^2

Speed,v =  32.8 m/s

To Find:

The coefficient of  static friction between the tires and the road = ?

Solution:

The radial Acceleration is given by,

Now the total acceleration is

=>

=>

=>

=>

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of  static friction is

From (1) and (2)

Substituting the values, we get

energy can be converted from one form to another, but it cannot be created or destroyed in ordinary chemical or physical means. do not understand what you are asking but hope this

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.