(a) 95.9 m
The initial velocity of the car is
[tex]u=53.0 km/h = 14.7 m/s[/tex]
The car moves by uniformly accelerated motion, so we can use the SUVAT equation:
[tex]v^2 - u^2 = 2ad[/tex]
where
v = 0 is the final velocity
d is the stopping distance of the car
a is the acceleration of the car
The force of friction against the car is
[tex]F_f = - \mu mg[/tex]
where
[tex]\mu=0.115[/tex] is the coefficient of friction
m is the mass of the car
[tex]g = 9.8 m/s^2[/tex] is the acceleration due to gravity
According to Newton's second law, the acceleration is
[tex]a=\frac{F}{m}=\frac{-\mu mg}{m}=-\mu g[/tex]
Substituting into the previous equation:
[tex]v^2 - u^2 = -2\mu g d[/tex]
and solving for d:
[tex]d=\frac{v^2 -u^2}{-2\mu g}=\frac{0-(14.7 m/s)^2}{-2(0.115)(9.8 m/s^2)}=95.9 m[/tex]
(b) 19.1 m
This time, the coefficient of friction is
[tex]\mu = 0.575[/tex]
So the acceleration due to friction is:
[tex]a=-\mu g = -(0.575)(9.8 m/s^2)=-5.64 m/s^2[/tex]
And substituting into the SUVAT equation:
[tex]v^2 - u^2 = 2ad[/tex]
we can find the new stopping distance:
[tex]d=\frac{v^2 -u^2}{-2a}=\frac{0-(14.7 m/s)^2}{2(-5.64 m/s^2)}=19.1 m[/tex]
