Answer:
The minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Explanation:
We know by equation of motion that,
![v^{2}=u^{2}+2\cdot a \cdot s](https://tex.z-dn.net/?f=v%5E%7B2%7D%3Du%5E%7B2%7D%2B2%5Ccdot%20a%20%5Ccdot%20s)
Where, v= final velocity m/sec
u=initial velocity m/sec
a=Acceleration m/![Sec^{2}](https://tex.z-dn.net/?f=Sec%5E%7B2%7D)
s= Distance traveled before stop m
Case 1
u= 13 m/sec, v=0, s= 57.46 m, a=?
![0^{2} = 13^{2} + 2 \cdot a \cdot57.46](https://tex.z-dn.net/?f=0%5E%7B2%7D%20%3D%2013%5E%7B2%7D%20%20%2B%202%20%5Ccdot%20a%20%5Ccdot57.46)
a = -1.47 m/
(a is negative since final velocity is less then initial velocity)
Case 2
u=29 m/sec, v=0, s= ?, a=-1.47 m/
(since same friction force is applied)
![v^{2} = 29^{2} - 2 \cdot 1.47 \cdot S](https://tex.z-dn.net/?f=v%5E%7B2%7D%20%3D%2029%5E%7B2%7D%20%20-%202%20%5Ccdot%201.47%20%5Ccdot%20S)
s = 285.94 m
Hence the minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Never is the correct answer
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It will be stand 46.67 all i did was divide both numbers but im not sure if im right but i hope i am hope i helped:)
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