Question

A car is traveling at 60.6 mph 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 0.056, what is the minimum distance in which the car will stop? Answer in units of m. m = 60-kg.

Answer 1

for the min distance condition : you must be knowing that maximum/limiting friction is acting. It'll provide a deceleration of (coeff)*g . Please convey any doubts.

Answer 2

you want direct answer or procedure

Answer 3

IF the friction coefficient stays constant during braking then:
60,6mph=27m/s
B - friction coefficient
F(friction)=B*N, where N is the normal reaction, I assume that the car is on horisontal road, so N=m*g, where m is the mass of the car.
F(braking)=m*a, where a is the acceleration.
On a limiting case, where F(braking)=F(friction), we get a=B*g.
From the equation x=v^2/2*a (because v(final)=0), we get x=(v^2)/(2*B*g)=660m
In conclusion, this friction coefficient is ridiculously low, hence the long distance to stop.

Answer 4

what is B?

Answer 5

he meant B = friction coefficient

Answer 6

Ok thank you @kappa007

Answer 7

:)