# Solving for the friction without mass

## Homework Statement

A car is traveling at 48.0 km/h on a flat highway.
(a) If the coefficient of friction between road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop?

(b) What is the stopping distance when the surface is dry and the coefficient of friction is 0.600?

## Homework Equations

F = ma
Ff = u*Fn
kinematics equations

## The Attempt at a Solution

I attempted to get the F in F = ma to equal the friction equation because friction is the only force acting on the car.
so i got:
F = Ff
ma = μ*Fn
ma = μ*m*g
a = μ*g
a = .1*-9.8
a = -.98

Vf^2 = Vi^2 + 2aX
0 = 48^2 + 2(-.98)(X)
-2304 = -1.96 (X)
1175.5 = X

But this isn't correct... So i was hoping someone would be able to show me the error of my ways so i could fix part a and do part b

SteamKing
Staff Emeritus
Homework Helper
Check your units. Do you think it is reasonable for a car travelling at a moderate speed to take over 1000 meters to stop?

it doesnt seem reasonable...

could i get a hint potentially?

ehild
Homework Helper
What is the unit of your x?

ehild

meters

ehild
Homework Helper
Why? Substitute the data together with their units.

ehild

Why? Substitute the data together with their units.

ehild

What?

ehild
Homework Helper
You calculated x from the equation vf^2=vi^2-2ax. What are the units of vf, vi and a?

ehild

v = m/s
g = m/s/s

v = km/h

ehild
Homework Helper
Do you get the right values when substituting v is m/s?

ehild

could you explain that a little more...?

ehild
Homework Helper
X=vi^2/(2a). If you substitute vi in km/h, you get x in (km/h)^2s^2/m. Does it have sense? You need to convert km/h to m/s to get X in meters.

$$X=\frac{(V (m/s))^2}{2*0.1*9.8 (m/s^2)}=\frac{(V (m/s))^2}{2*0.1*9.8 (m/s^2)}$$

ehild

X=vi^2/(2a)

i used this equation to get 1175.51 km.
to get meters i would divide by 1000, right?
to get a stopping distance of 1.17551 meters?

SteamKing
Staff Emeritus
Homework Helper
The car is travelling 48 km/hr. Are these units consistent with a g = 9.81 m/s/s?

I got it!
i wasn't doing the conversion right.
right away when the velocity is 48 km/hr i needed to change that to meter/sec. so 48 km/hr = 13.3 m/s.
then plug 13.3 into the equation X=vi^2/(2μa)
resulting in
A = 90.25 m
B = 15.041 m
THANK YOU VERY MUCH!!!

they are not consistent...
soo i would convert 48 km/hr to 13.3 m/s
so using x=v^2/(2ua)
for (a) i got 90.25
and (b) i got 15.041

These seem accurate... thank you very much for your help!!!!

ehild
Homework Helper
Good work at last!

ehild