what is the minimum distance in which the car will stop in units of meters?

Question

a car is traveling at 59.6 km/h on a flat highway. the acceleration of gravity is 9.8 m/s squared. the coefficient of friction between the road and the tires is .172.

Answer 1

For minimum distance, the friction force times stopping distance equals the kinetic energy.  (Note:  m/sec = km/hr / 3.6)

m*(59.6/3.6)²/2 = m*d*9.8*0.172

m drops out, yielding:

d*9.8*.172 = (59.6/3.6)²/2

d = (59.6/3.6)²/(2*9.8*0.172)

d = 81.3 meters

Answer 2

The stopping force would be mgk, where m = mass of car, g = gravity acceleration and k = coefficient of friction. Using F = ma and solving for a we get the deacceleration a = F/m , so combining this equation and the first equation we get a = gk, fortunately the mass drops out since we don't know it. So the car will deaccelerate at (9.8)(.172) = 1.69m/s^2 since 59.6km/h = 16.55 m/sec You can now find the time by dividing the speed 16.55/1.69 = 9.79 seconds

Answer 3

Ok, what u need to to is: Calculate the friction force (Fr), which holds the car back, than calcualte the time (Ts) it takes the friction force to reduce the speed of the car to zero. Than calculate the distance (Xs) the car moves while deceleration.

So:
Fr = 9.81 * 0.172 = 1.68 N
59.6 Km/h are 16.555 m/s (59.6 / 3.6), and we always work with m/s.

Ts=16.555/1.68 = 9.81s , so to car needs 9.81 seconds to stop.

Now use the standart "moving" function: X(t)= 1/2 * a * t^2 + v *t + X0.

Xs= X(t=9.81)= -1.68 / 2 * 9.81^2 + 16.555 * 9.81 + 0 = 81.21

So it takes the car 9.81 seconds , and 81.21 meters to stop.