A point moves with deceleration along a circle of radius $R$ such that at any moment of time, the ma

Question

A point moves with deceleration along a circle of radius $R$ such that at any moment of time, the magnitudes of its tangential and normal accelerations are equal. At the initial moment $t = 0$, the velocity of the point is $v_0$. Which of the following statements is/are CORRECT?

Professor PC Thomas Classes · Answer

Decelerating $\implies w_	au = -dv/dt = -v(dv/ds)$. Given $|w_	au| = w_n = v^2/R$. So $-dv/dt = v^2/R \implies -dv/v^2 = dt/R \implies 1/v - 1/v_0 = t/R \implies v(t) = \frac{v_0 R}{R + v_0 t}$ (A is correct). Also $-v(dv/ds) = v^2/R \implies -dv/v = ds/R \implies \ln(v/v_0) = -s/R \implies v(s) = v_0 e^{-s/R}$ (B is correct). Total acceleration $w = \sqrt{w_n^2 + w_	au^2} = \sqrt{2}(v^2/R) = \frac{\sqrt{2}v_0^2}{R}e^{-2s/R}$ (C is correct). Velocity never reaches 0 for finite $s$ (D is incorrect).