# static friction problem without mass

I am having trouble with a physics problem. The problem is as follows:

A car is traveling at 54.0 mi/h on a horizontal highway.

A) If the coefficient of static friction between road and tires on a rainy day is 0.103, what is the minimum distance in which the car will stop?

B) What is the stopping distance when the surface is dry and µs = 0.595?

This is what I have so far for A)

fs = µs * Fn

Fn = Mg
fs = Ma

Ma = µs * Mg
a = µs * g

V = 54 - a * t
0 = 54 - a * t
t = 53.497 seconds

dist = 54*t - (a/2)*t^2
= 1444.42243 miles
= 2324572.57 meters


This is not the correct answer. Any help would me much appreciated.

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Hi Patrick, and welcome to Physics Stack Exchange! What specifically do you think might be wrong with what you've done? – David Z Feb 21 '12 at 2:13
It turns out my calculations are correct, but I didn't fully take units into consideration. As Akhmenteli pointed out, the cars speed is given in mi/hr and my answer is m/hr – Patrick Lorio Feb 21 '12 at 2:49
OK, well just something to keep in mind for the future: we much prefer it if you ask a conceptual question (such as asking about something specific you think you might be doing wrong), don't just ask for people to check your work. – David Z Feb 21 '12 at 2:55

## 2 Answers

The first issue I see: your speed is in miles/hour, and your time is in seconds.

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that might be it, thanks. – Patrick Lorio Feb 21 '12 at 2:34

$$a=\mu g$$

$u=86.4 \:\mathrm{km/hr}$

at stoppage,

$$v=0$$

we know that

$$v^2-u^2=2as$$

$$0-(86.4*86.4)=-2*0.103*10*\text{distance}$$

$$7464.96/2.06=\text{distance}$$

$$3623.76\:\mathrm{km}=\text{distance}$$

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