Answer :
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]
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:
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