Question

A particle moves with deceleration along the circle of radius $R$ so that at any moment of time its tangential and centripetal 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. $v=v_0{e}^{-R/s}$
• D. $v=v_0{e}^{R/s}$

Answer 1

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

$\left|\frac{dv}{dt}\right|=\frac{v^2}{R}$
$\Rightarrow -\frac{dv}{dt}=\frac{v^2}{R}$
$\Rightarrow -\frac{dv}{ds}\times \frac{ds}{dt}=\frac{v^2}{R}$
$\Rightarrow \frac{-dv}{ds}v=\frac{v^2}{R}$
$\Rightarrow {\int }_{v_0}^v-\frac{dv}{v}={\int }_0^s\frac{ds}{R}$
$\Rightarrow \ln \frac{v}{v_0}=-\frac{s}{R}$
$\Rightarrow \frac{v}{v_0}=e^{-s/R}$
$\Rightarrow v=v_0{e}^{-s/R}$