Asked by grace
A car is travelling at 42.4 km/h on a flat highway. The acceleration of gravity is 9.81 m/s^2. If the coefficient of friction between the road and the tires on a rainy day is 0.125 what is the minimum distance for the car to stop? Answer is metres
Answers
Answered by
drwls
First, convert speed (V) to units of m/s.
42.4 km/h = 11.78 m/s
Initial kinetic energy = Work done against friction
(1/2)MV^2 = M*g*Uk*X
M cancels out.
X = V^2/(2*g*Uk) is the stopping distance, if the car is skidding.
If the brakes manage to apply the maximum force that prevents a skid as an automatc braking system (ABS) is supposed to do, the static friction coefficient Us can be used instead of Uk, the kinetic friction coefficient. In that case you get a minimum stopping distance.
They don't say which friction coefficient they are providing.
42.4 km/h = 11.78 m/s
Initial kinetic energy = Work done against friction
(1/2)MV^2 = M*g*Uk*X
M cancels out.
X = V^2/(2*g*Uk) is the stopping distance, if the car is skidding.
If the brakes manage to apply the maximum force that prevents a skid as an automatc braking system (ABS) is supposed to do, the static friction coefficient Us can be used instead of Uk, the kinetic friction coefficient. In that case you get a minimum stopping distance.
They don't say which friction coefficient they are providing.
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