Question

Starting from rest a particle moves upon a circular path of radius R such that the magnitude of tangential acceleration is inversely proportional to speed (v) then​

Answer 1

Given:

Starting from rest a particle moves upon a circular path of radius R such that the magnitude of tangential acceleration is inversely proportional to speed (v).

To find:

Relation between velocity, distance and normal acceleration ?

Calculation:

As per the question :

$\therefore \: a_{t} \propto \dfrac{1}{v}$

$\implies\: a_{t} = \dfrac{k}{v} \: . \: . \: . \: .(k \: is \: a \: constant)$

$\implies\: \dfrac{dv}{dt} = \dfrac{k}{v}$

$\implies\: v \: dv = k \: dt$

$\displaystyle\implies\: \int_{0}^{v} v \: dv = k \: \int_{0}^{t} dt$

$\implies \: \dfrac{ {v}^{2} }{2} = kt$

$\implies \: {v}^{2} =2 kt$

$\implies \: v = \sqrt{2 kt}$

$\implies \: v = c \sqrt{t} \: . \: . \: . \: .(c = \sqrt{2k} )$

$\implies \: \dfrac{ds}{dt} = c \sqrt{t} \: . \: . \: . \: .(c = \sqrt{2k} )$

$\implies \: ds= c \sqrt{t} \: dt$

$\displaystyle \implies \: \int_{0}^{x} ds= c \int_{0}^{t}\sqrt{t} \: dt$

$\implies \: s = \dfrac{2c}{3} \times {t}^{ \frac{3}{2} }$

$\implies \: s \propto {t}^{ \frac{3}{2} }$

$\implies \: t \propto \: {s}^{ \frac{2}{3} }$

$\implies \: {v}^{2} \propto \: {s}^{ \frac{2}{3} }$

$\boxed{ \implies \: v \propto \: {s}^{ \frac{1}{3} } }$

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$\therefore \: a_{N} = \dfrac{ {v}^{2}}{R}$

$\implies \: a_{N} \propto {v}^{2}$

$\implies \: a_{N} \propto {( \sqrt{t} )}^{2}$

$\boxed{\implies \: a_{N} \propto t}$

So, option 4 is correct.