- #1
kza62
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A car is traveling at 40.7 mi/h on a horizontal highway.
The acceleration of gravity is 9.8 m/s/s.
If the coefficient of friction between road and tires on a rainy day is 0.12, what is the minimum distance in which the car will stop?
(1 mi = 1.609 km)
So given:
vi(initial velocity): 40.7 mi/h -> 65486.3m/3600s
vf(final velocity): 0 m/s
g(gravity): -9.8
Ff(friction): -0.12
m(mass): ?
Fnet(net force): ?
x(distance): ?
Equations:
F = ma
Fnet = ma - Ff
Ff = uFn
Fn = mg
vf(^2) = vi(^2) + 2ax
Answer in back of textbook: 140.689 m
I am unsure how to set this problem up. All my attempts have had no success. Please, help would be appreciated.
The acceleration of gravity is 9.8 m/s/s.
If the coefficient of friction between road and tires on a rainy day is 0.12, what is the minimum distance in which the car will stop?
(1 mi = 1.609 km)
So given:
vi(initial velocity): 40.7 mi/h -> 65486.3m/3600s
vf(final velocity): 0 m/s
g(gravity): -9.8
Ff(friction): -0.12
m(mass): ?
Fnet(net force): ?
x(distance): ?
Equations:
F = ma
Fnet = ma - Ff
Ff = uFn
Fn = mg
vf(^2) = vi(^2) + 2ax
Answer in back of textbook: 140.689 m
I am unsure how to set this problem up. All my attempts have had no success. Please, help would be appreciated.