Solving for the friction without mass

[ngorecki]

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]

Check your units. Do you think it is reasonable for a car traveling at a moderate speed to take over 1000 meters to stop?

 

[ngorecki]

it doesn't seem reasonable...

 

[ngorecki]

could i get a hint potentially?

 

[ehild]

What is the unit of your x?ehild

 

[ngorecki]

meters

 

[ehild]

Why? Substitute the data together with their units.

ehild

 

[ngorecki]

Why? Substitute the data together with their units.

ehild

What?

 

[ehild]

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

ehild

 

[ngorecki]

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

 

[ngorecki]

v = km/h
my bad

 

[ehild]

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

ehild

 

[ngorecki]

could you explain that a little more...?

 

[ehild]

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

 

[ngorecki]

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]

The car is traveling 48 km/hr. Are these units consistent with a g = 9.81 m/s/s?

 

[ngorecki]

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!

 

[ngorecki]

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]

Good work at last!

ehild