1. Feb 7, 2006

### Jacob87411

A car is traveling at 46.0 mi/h on a horizontal highway.
If the coefficient of static friction between road and tires on a rainy day is 0.101, what is the minimum distance in which the car will stop?

Just confused on where to start becuase it seems like I need more info

2. Feb 7, 2006

### marlon

There is something missing. Don't you have a weight of the car ?

You should apply :
$$x = v_ot -a \frac{t^2}{2}$$
$$v = v_o -at$$

Where the v_o is initial velocity and a is the acceleration due to the friction. Beware of the sign !!!

Solve the second equation for t (the v is zero when the car has stopped) and substitute this t in the first equation to solve it for x.

marlon

3. Feb 7, 2006

### Jacob87411

Thats what I thought but there is no mass given

4. Feb 7, 2006

### marlon

ohh yes, just apply newton's second law in the vertical Y-direction :

$$ma_y = 0 = N -mg$$

normal force N equals N = mg

and friction force F = 0.101N = 0.101mg

The associated acceleration is F/m = 0.101g and g=9.81 m/s²

marlon