A car is traveling at 40.7 mi/h on a horizontal highway.(adsbygoogle = window.adsbygoogle || []).push({});

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 text book: 140.689 m

I am unsure how to set this problem up. All my attempts have had no success. Please, help would be appreciated.

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# Help! Forces and Friction and Movement! O my!

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