a) 52.5 m
b) 16.0 m/s
Explanation:
a)
The motion of a car slowed down by friction is a uniformly accelerated motion, so we can use the following suvat equation:
[tex]v^2-u^2=2as[/tex]
where
v = 0 is the final velocity (the car comes to a stop)
u = 28.7 m/s is the initial velocity of the car
a is the acceleration
s is the stopping distance
For a car acted upon the force of friction, the acceleration is given by the ratio between the force of friction and the mass of the car, so:
[tex]a=\frac{-\mu mg}{m}=-\mu g[/tex]
where:
[tex]\mu=0.80[/tex] is the coefficient of friction
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
Substituting and solving for s, we find:
[tex]s=\frac{v^2-u^2}{-2\mu g}=\frac{0-(28.7)^2}{-2(0.80)(9.8)}=52.5 m[/tex]
b)
In this case, the car is moving on a wet road. Therefore, the coefficient of kinetic friction is
[tex]\mu=0.25[/tex]
Here we want the stopping distance of the car to remain the same as part a), so
[tex]s=52.5 m[/tex]
We can use again the same suvat equation:
[tex]v^2-u^2=2as[/tex]
And since the final velocity is zero
u = 0
We can find the initial velocity of the car:
[tex]v=\sqrt{2as}=\sqrt{2\mu gs}=\sqrt{2(0.25)(9.8)(52.5)}=16.0 m/s[/tex]
