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Chapter 04, Problem 063 At t, -5.00 s, the acceleration of a particle moving at constant speed in counterclockwise dircular motion is a1 = (6.00m/s*)i + ( 10.0m/s*)/I At t2 7.00 s (less than one period later), the acceleration is #2 = ( 10.0m/s2)/-(6.00m/s2)/ The period is more than 2.00 s. What is the radius of the cirde? Units Number the tolerance is +/-2% Click if you would like to Show Work for this question: Open Sho Work
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Answer #1

a = v^2/r

a = sqrt(10^2 + 6^2) = sqrt(136)

Distance particle has moved = (angle moved through)*radius = (pi/2)*r
Distance = velocity*time = v*t
v*t = (pi/2)*r
Using t = 2 we have v = (pi/4)*r

a = [(pi/4)*r]^2
r = a/[pi^2/16] = 16*a/pi^2 = 16*sqrt(136)/pi^2
r = 18.90 meters

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