Question

A particle 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 speed of the particle equals $v_0$, then the speed of the particle as a function of the distance covered $S$ will be

• A. $v=v_0{e}^{-\frac{S}{R}}$
• B. $v=v_0{e}^{-\frac{S}{R}}$
• C. $v=v_0{e}^{-\frac{R}{S}}$
• D. $v=v_0{e}^{\frac{R}{S}}$

Answer 1

The correct option is A $v=v_0{e}^{-\frac{S}{R}}$

Given that magnitude of centripetal and tangential acceleration are equal, so
$\frac{dv}{dt}=-\frac{v^2}{R}\Rightarrow \frac{dv}{ds}=-\frac{v}{R};{\int }_{v_0}^v\frac{-1}{v}dv={\int }_0^S\frac{ds}{R}$
$\Rightarrow ln\left[\frac{v_0}{v}\right]=\frac{S}{R}\Rightarrow \frac{v_0}{v}=e^{\frac{S}{R}}$
$\Rightarrow v_0=v{e}^{\frac{S}{R}}\Rightarrow v=v_0{e}^{-\frac{S}{R}}$