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Sagot :
Given data
*The given coefficient of static friction between the coeffee cup and the roof of the car is
[tex]\mu=0.24[/tex]*The value of the acceleration due to gravity is g = 9.8 m/s^2
*The given initial speed of the car is u = 0 m/s
*The given final speed of the car is v = 15 m/s
(a)
The formula for the maximum acceleration of the car can have without causing the cup to slide is given as
[tex]a=\mu g[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} a=(0.24)(9.8) \\ =2.35m/s^2 \end{gathered}[/tex]Hence, the maximum acceleration of the car can have without causing the cup to slide is a = 2.35 m/s^2
(b)
The formula for the smallest amount of time in which the person can accelerate the car from rest to 15 m/s is given by the equation of motion as
[tex]\begin{gathered} v=u+at \\ t=\frac{v-u}{a} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} t=\frac{15-0}{2.35} \\ =6.37\text{ s} \end{gathered}[/tex]Hence, the smallest amount of time in which the person can accelerate the car from rest to 15 m/s is t = 6.37 s
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