# Percent Length Contraction (check solution)

by Quelsita
Tags: check, contraction, length, percent, solution
 P: 50 Question: What is the percent length contraction of an aircraft traveling at Mach 2? So, we know that Mach 2= 680.58 m/s and that L'=L$$\sqrt{1-(v/c)^2}$$ If you divide over the L to get: L'/L=$$\sqrt{1-(v/c)^2}$$=% length contraction Plug-n-chug from here to get: L'/L=$$\sqrt{1-(680.58/c)^2}$$ =$$\sqrt{1-(5.15X10^-12)}$$ =$$\sqrt{1}$$ =1 Is this correct? It only contracts 1%?
 HW Helper P: 5,346 I'd recheck your calculation and be careful in taking your square root. 680/300,000 is what you were intending I trust?
P: 50
 Quote by LowlyPion 680/300,000 is what you were intending I trust?
Yep. That's why I thought it was off because it's such a small number that you'll have 1 under the radical...

Or is the percent contraction supposed to be very small since the aircraft, compared to the speed of light, is going extremely slow?

HW Helper
P: 5,346

## Percent Length Contraction (check solution)

Yes it is a small number.

And when you take the square root it gets closer to 1.

Use a calculator, and don't approximate or round until you have an expression for the percentage.
P: 181
 Quote by LowlyPion I'd recheck your calculation and be careful in taking your square root. 680/300,000 is what you were intending I trust?
c = 300,000,000 m/s

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