Asked by Elizabeth
A car is traveling at 54.2 km/h on a flat highway. The acceleration of gravity is 9.81 m/s^2.
PART 1: If the coefficient of kinetic friction between the road and the tires on a rainy day is 0.131, what is the minimum distance needed for the car to stop? Answer in unitsof m.
PART 2: What is the stopping distance when the surface is dry and the coefficient of kinetic friction is 0.603? Answer in units of m.
PART 1: If the coefficient of kinetic friction between the road and the tires on a rainy day is 0.131, what is the minimum distance needed for the car to stop? Answer in unitsof m.
PART 2: What is the stopping distance when the surface is dry and the coefficient of kinetic friction is 0.603? Answer in units of m.
Answers
Answered by
MathMate
Frictional resistance, F
= -μ mg
acceleration, a
= F/m
= -μg
For distance S, use
v²-u²=2aS
where v=final velocity =0
u=initial velocity = 54.2 km/h = 15.06 m/s
Solve for S.
1. μ=0.131
u=54.2
a=-0.131*9.8=-1.284 m/s²
S=15.06²/(-2*1.284)
=88.3m
The same approach can be used for both cases.
Post your answers for a check if you wish.
= -μ mg
acceleration, a
= F/m
= -μg
For distance S, use
v²-u²=2aS
where v=final velocity =0
u=initial velocity = 54.2 km/h = 15.06 m/s
Solve for S.
1. μ=0.131
u=54.2
a=-0.131*9.8=-1.284 m/s²
S=15.06²/(-2*1.284)
=88.3m
The same approach can be used for both cases.
Post your answers for a check if you wish.
Answered by
Elizabeth
omg thx so much!!!!