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Ke Min is driving at 23 m/s in a 1200-kg car. He sees a tree branch lying across the road. He slams on the brakes when the branch is 60.0 m in front of him. If the coefficient of kinetic friction between the car’s locked tires and the road is 0.41, how much distance is required for the car to come to a stop?

Sagot :

Answer:

x = 65.82877053 m

Explanation:

Fnet = -F f  = ma

-μ Fnorm = ma

(-μ) mg  = ma

(-μ) (g) = a

[tex]a = -(mu) g = -(0.41)(9.80 m/s^2) = -4.018 m/s^2\\\\v^2 = vinitial^2 + 2ax\\x = \frac{v^2 - vinitial^2}{2a} = \frac{0 - (23m/s)^2}{2(-4.018 m/s^2}\\x = 65.82877053 m[/tex]

snehashish65

The distance required to stop by car is 65.96 m.

Given data:

The initial speed of Ke Min is, u = 23 m/s.

The mass of car is, m = 1200 kg.

The distance between the branch and Ke Min is, d = 60.0 m.

The coefficient of kinetic friction between the car’s locked tires and the road is, [tex]\mu =0.41[/tex].

Here in the given problem, the frictional force will oppose the motion of the car. So, the linear force causing the acceleration is,

F = f

[tex]ma = -\mu \times m \times g\\\\a = -\mu \times g\\\\a = -0.41 \times 9.8\\\\a =-4.01 \;\rm m/s^{2}[/tex]

Now, apply the third kinematic equation of motion to obtain the stopping distance (s) as,

[tex]v^{2} = u^{2} +2as\\\\0^{2} = 23^{2} +2(-4.01)s\\\\s = \dfrac{529}{8.02}\\\\s =65.96 \;\rm m[/tex]

Thus, we can conclude that the distance required to stop by car is 65.96 m.

Learn more about the kinematic equations of motion here:

https://brainly.com/question/14355103

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