Displacement given coefficient of friction and velocity

AI Thread Summary
A car traveling at 56.0 km/h on a flat highway with a coefficient of friction of 0.100 will stop in a minimum distance of 245.3 meters. The acceleration due to friction is calculated as 0.98 m/s² using the formula μ = -a/g. The initial velocity is converted to 15.6 m/s, which is then used in the kinematic equation v² = v₀² + 2ax to find the stopping distance. There was a correction needed in the math regarding the signs in the calculations. Understanding the relationship between the coefficient of friction and acceleration is crucial for solving such problems.
mandy9008
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Homework Statement


A car is traveling at 56.0 km/h on a flat highway. If the coefficient of friction between road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop?



Homework Equations


μ=-a/g
v2=vo2 + 2 a x


The Attempt at a Solution


56.0 km/h = 15.6 m/s

0.100=-a / 9.8 m/s2
a= 0.98 m/s2

0=(15.6 m/s)2 + 2 (0.98m/s2) x
x= 245.3 m
 
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mandy9008 said:
0=(15.6 m/s)2 + 2 (0.98m/s2) x
x= 245.3 m
Your equation is good but check your math(s)...and minus sign
 
oh i accidently subtracted instead of divided. oops thanks!
 
How do you no the equation:
the coefficient of friction=-a/g ?
 

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