(a) [tex]2.2 m/s^2[/tex]
Along the vertical direction, the acceleration of the cup is zero, therefore the equation of the forces is:
[tex]N-mg=0[/tex]
where N is the normal reaction and (mg) is the weight of the cup, with m being its mass and g the acceleration of gravity. From this equation we get
[tex]N=mg[/tex] (1)
Along the horizontal direction, the only force acting is the force of friction. The maximum force of friction acting on the cup is
[tex]f_f = \mu N[/tex]
where [tex]\mu=0.22[/tex] is the coefficient of static friction between the cup and the roof of the car. The maximum acceleration that the car can substain withouth the cup sliding away corresponds to the maximum force of friction on the cup, so we can write:
[tex]F =\mu N = ma_{max}[/tex]
And solving for [tex]a_{max}[/tex] and using eq.(1), we get:
[tex]a_{max} = \mu g = (0.22)(9.8)=2.2 m/s^2[/tex]
(b) 5.9 s
To solve this part of the problem, we can use the following SUVAT equation:
[tex]v = u+at[/tex]
where
v is the final velocity
u is the initial velocity
a is the maximum acceleration
t is the smallest amount of time
In this problem,
v = 13 m/s (final velocity of the car)
u = 0 (initial velocity)
a = 2.2 m/s^2
Solving for t,
[tex]t=\frac{v-u}{a}=\frac{13}{2.2}=5.9 s[/tex]
