Question

A body moves in a circular path of the radius $R$ with deceleration so that at any moment of time its tangential and normal acceleration are equal in magnitude. At the initial moment $t=0$ , the velocity of the body is $v_0$ then the velocity of the body after it has moved $S$ at any time will be -

• A. $v=v_0{e}^{-\frac{2S}{R}}$
• B. $v=v_0{e}^{-\frac{S}{R}}$
• C. $v=v_0{e}^{-SR}$
• D. $v=v_0{e}^{-2SR}$

Answer 1

Answer: (B)

$a_T=a_N$
$\frac{vdv}{ds}=\frac{-v^2}{R}$
(-ve because it is a retardation) and $s$ is the distance. covered
tangentially on circular path
on integrating we get
${\int }_{v_0}^v\frac{dv}{v}={\int }_0^s\frac{-ds}{R}$
${\left[{\log }_V\right]}_{v_0}^v=\frac{1}{R}(-1)s$
$\log \left(\frac{v}{v_0}\right)=\frac{-s}{R}$
$V=v_0{e}^{-\frac{s}{R}}$