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{2 S}{R}} \, $
• B. $v=v_{0}e^{- \frac{S}{R}}$
• C. $v=v_{0}e^{- S R}$
• D. $v=v_{0}e^{- 2 S R}$

Answer 1

Answer: (B)

$a_{T}=a_{N}$
$\frac{v d v}{d s}=\frac{-v^{2}}{R}$
(-ve because it is a retardation) and $s$ is the distance. covered
tangentially on circular path
on integrating we get
$\int_{ v _{0}}^{ v } \frac{ dv }{ v }=\int_{0}^{ s } \frac{- ds }{ R }$
$\left[\log _{ V }\right]_{ v _{0}}^{ v }=\frac{1}{ R }(-1) s$
$\log \left(\frac{ v }{ v _{0}}\right)=\frac{- s }{ R }$
$V = v _{0} e ^{-\frac{ s }{ R }}$