First of all, let's convert the initial speed from mi/h to m/s.
We know that 1 mi=1609 m and 1 h=3600 s. So, the conversion factor is:
[tex]1 \frac{mi}{h}= 1 \frac{1609 m}{3600 s}=0.45 [/tex]
And the initial speed of the car is therefore:
[tex]v_i = 52 mi/h \cdot 0.45 = 23.4 m/s[/tex]
Now we can start to solve the problem.
a) I assume that 0.102 is the coefficient of dynamic friction, not static friction, because the car is in motion.
The only force acting on the car is the frictional force, directed against the motion of the car. We can write Newton's second law for the car:
[tex]ma=-\mu mg[/tex]
where m is the mass of the car, a its acceleration, and [tex]\mu m g[/tex] is the frictional force, with a negative sign since it is acting against the motion of the car. Simplifying m and solving, we can find the acceleration of the car:
[tex]a=-\mu g=-(0.102)(9.81 m/s^2)=-1 m/s^2[/tex]
The acceleration is negative because the car is decelerating.
Now we can use the following relationship valid for a uniformly accelerated motion:
[tex]2aS = v_f^2-v_i^2[/tex]
where S is the distance covered, vi and vf are the initial and final speed of the car. Since the car is going to stop, [tex]v_f=0[/tex], and by replacing a and viwith the values we calculated previously, we can find the distance S that the car needs to stop:
[tex]S=- \frac{(v_i)^2}{2a}= -\frac{(23.4 m/s)^2}{2(-1 m/s^2)}=273.8 m [/tex]
b) The procedure to solve this part of the exercise is exactly identical to the first part, the only difference is the value of the coefficient of friction: [tex]\mu=0.603[/tex]. Let's start by calculating the new acceleration
[tex]a=-\mu g=-(0.603)(9.81 m/s^2)=-5.9 m/s^2[/tex]
And then, the new distance needed to stop the car is
[tex]S=- \frac{(v_i)^2}{2a}=- \frac{(23.4 m/s)^2}{2(-5.9 m/s^2)}=46.4 m [/tex]
So, when the surface is dry the car needs much less space to stop completely than when the surface is wet.
