At tl = 2.00 s, the acceleration of a particle in

Chapter , Problem 63

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At \(t_{1}=2.00 \mathrm{~s}\), the acceleration of a particle in counterclockwise circular motion is \(\left(6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(4.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{j}\). It moves at constant speed. At time \(t_{2}=5.00 \mathrm{~s}\), the particle’s acceleration is \(\left(4.00 \mathrm{~m} / \mathrm{s}^{2}\right) \mathrm{i}+\left(-6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{j}\). What is the radius of the path taken by the particle if \(t_{2}-t_{1}\) is less than one period?

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