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Kinematics

Problem 1.38

A point moves with deceleration along the circle of radius \(R\) so that at any moment of time its tangential and normal accelerations are equal in moduli. At the initial moment \(t=0\) the velocity of the point equals \(v_{0} .\) Find: (a) the velocity of the point as a function of time and as a function of the distance covered \(s\); (b) the total acceleration of the point as a function of velocity and the distance covered.

Reveal Answer
 (a) v=v0/(1+v0t/R)=vne−z/R (b) w=2‾v02/Re2s/R==2‾v2/R\text { (a) } v=v_{0} /\left(1+v_{0} t / R\right)=v_{n} \mathrm{e}^{-z / R} \text { (b) } w=\sqrt{\overline{2}} v_{0}^{2} / R \mathrm{e}^{2 s / R}= =\sqrt{\overline{2}} v^{2} / R
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