Arc Length Units: Explained & Solved Problem

alingy2
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Hello,

I solved the arc length for a particular problem. However, what is the unit of arc length if the units of the velocity vs time graph are m/s vs s?

I am really confused.
 
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Well, suppose you wanted to know the arc length of a portion of a circle; the typical formula is arc length = radius*angle. So the units would be meters*radians in standard SI. Although radians is a dimensionless unit, so you could also say units are just meters.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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