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 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.

 

[tiny-tim]

Welcome to PF!

Hi kza62! Welcome to PF!

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

i] what is the normal force?

ii] what is the friction force?

iii] what is the acceleration?

 

[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 !

(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 µF_n, so what you found were the accelerations, but since the two masses cancel, it makes no difference )

 

[kza62]

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