Solving Uncertainty Problem with .7g Lateral Acceleration

[purduegrad]

Here's the problem:

A magazine publishes lateral acceleration capability from cars it tests. Measurements are made using a 150' diameter skidpad, the vehicle path deviates from the circle +/- 2 ft and the vehicle speed is read from a fifth wheel sensor measuring the system to +/- .5 mph. Then it says to estimate the experimental uncertainty if the reported lateral acceleration is .7g.

All I can think of is A=v^2/r...and some how i would have to cut the 150 diameter into 75 as a radius. I got 75 +/- .0133. then i found v = to 22.69, and you have to square that...I'm basically stumped...the answer is like 4.XX%

 

[Nenad]

I did the calculation and I got 8%.

 

[purduegrad]

hmm well its supposedly like 4.45%, i have a feeling this has something to do with partial derivatives?

 

[Nenad]

no, you don't need partial derivatives to solve it. Ill get back to you with the answer, by the way, how many feet in a mile?

 

[purduegrad]

5280 feet in mile

 

[Nenad]

I Still can't get it, I get something liek 3.8%.

 

[Chronos]

See if this helps
$$A = \frac{1.226r}{t^2}$$
where
A = acceleration in g's
r = radius of track
t = time [in seconds] to complete 1 lap
This is how g's are calculated on a skidpad
re: http://www.carcraft.com/howto/53698/index7.html

 

[purduegrad]

yeah but I am not given any information about times

 

[Chronos]

try substitution. T = d/v. The equation then becomes
$$A = \frac{1.226rv^2}{d^2}$$ Does that help?
Given the uncertainty of measurements in the quantities r and v, the answer is there. I agree with the 4.xxx% result, see what you get.

 

[purduegrad]

I'm still not following how to get that 4.45 answer, can you please elaborate for me?