I have a maths question asking the total % uncertainty in the velocity of a skidding car given by this equation provided by my teacher:
1/2 * m * v^2 = force * distance = m * g * f * d
where m = mass with an uncertainty of +/- 4%
g = 9.80665 m/s^2
f = coefficient of friction with an uncertainty of +/- 5%
d = length of skid marks +/- 3%
I simplified the equation, removing m:
v^2 = 2·g·f·d
Then I calculated total uncertainty of v^2 to be the uncertainty of f + the uncertainty of d = 5% + 3% = 8%
Three questions:
Am I correct in leaving out the constants 2 and g, since I am using relative uncertainties?
Should I have somehow also included the uncertainty of m, which I left out after simplifying the equation?
How would I then calculate the uncertainty of v from the uncertainty of v^2?
The rough answer is you are right in adding the uncertainties of $f$ and $d$. If they were both high instead of $fd$ you would have $1.05f1.03d=1.0815fd$. We ignore the quadratic part of the uncertainty, which is the $0.0015=0.05 \cdot 0.03$. You are also correct in dividing out the $m$ analytically and noting the resulting equation does not depend on $m$ so its uncertainty does not matter at all.
Once you have $\frac {\Delta (v^2)}{v^2}=8\%$ you can use the fact that $(1+a)^n \approx 1+an$ as long as $an \ll 1$. This gives $\frac {\Delta v}v = \frac 12\frac {\Delta (v^2)}{v^2}=4\%$