Help! Forces and Friction and Movement! O my!

[kza62]

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 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.

[tiny-tim]

Hi kza62! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)

i] what is the normal force?

ii] what is the friction force?

iii] what is the acceleration? :smile:

[kza62]

ok, i think i figured it out but i'm not sure how let me try to break it down again
so if Fgravity is -9.8 then the Fnormal is 9.8?
so is Ffriction = u * Fnormal?
which is 1.176 is a?
and then i used
vf - vi / 2a
which gave me 140.689

did i get it or was this a lucky guess?

[kza62]

*** vf^2 - vi^2 / 2a = x
and a = -1.176

[tiny-tim]

HI kza62kza62 ! :smile:

(just got up :zzz: …)
ok, i think i figured it out but i'm not sure how let me try to break it down again
so if Fgravity is -9.8 then the Fnormal is 9.8?
so is Ffriction = u * Fnormal?
which is 1.176 is a?
…
which gave me 140.689
*** vf^2 - vi^2 / 2a = x
and a = -1.176

Yes, that's fine!

(btw, the forces would be mass times g, or times µFn, so what you found were the accelerations, but since the two masses cancel, it makes no difference :wink:)

[kza62]

WOOHOO! thanx for ur help! i love physics when I get it :D