Two runners at the same point begin running in opposite directions along a circular track of radius $100$m at a speed of $5$m/s. At what rate is the (shortest) distance between them growing after $10$sec?
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Have you tried drawing a picture of where they are at $10$ seconds and which way they are going? Do you see any symmetries in this problem that will help you? Any other things you have tried? – David K Nov 05 '14 at 06:36
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The question is very close to this recent question. – André Nicolas Nov 05 '14 at 06:38
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Let's assume they travel with constant velocity $v$, and Radius of circle be $r$
After time $t$ Arc length will be $vt$ $$l=vt=r\theta$$ and angle subtending that arc will be $$\theta=\frac{vt}{r}$$ now shortest distance between them will be $$x=2r\sin\theta$$
Now we have our relation $$x=2r\sin\left(\frac{vt}{r}\right)$$ Now rate of change of shortest distance between them is $$\frac{dx}{dt}=2r\frac{v}{r}\cos\left(\frac{vt}{r}\right)=2v\cos\left(\frac{vt}{r}\right)$$ Now plug in values
Here's a rough diagram if it helps
Note: In figure after time $t$ their respective positions are shown by $A$ and $A'$
$$x=AA'$$ $$r=BA=BA'$$ $$\angle ABO=\angle A'BO=\theta$$
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Thank you, just a quick question about the solution. How did you get the "shortest distance formula"? (x=2rsin(vt/r)) – Bryce Nov 05 '14 at 06:59
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@Byce Actually It's simple trigonometry Hypotaneous of triangle is R then I've to find perpendicular side. – Nov 05 '14 at 07:02